RD Sharma Solutions for Class 11 Maths Chapter 33 Probability

RD Sharma Solutions Class 11 Maths Chapter 33 – Download Free PDF Updated for (2023-24)

RD Sharma Solutions for Class 11 Maths Chapter 33 – Probability are provided here for students to study and prepare for their board examination. In earlier classes, we have learnt about two approaches to the theory of probability, namely, (i) Statistical approach and (ii) Classical approach. Both theories have some serious deficiencies and limitations – let us discuss them in this chapter. RD Sharma Class 11 Maths Solutions contain detailed steps explaining all the problems that appear.

Chapter 33 – Probability contains four exercises, and RD Sharma Solutions provide comprehensive answers to all the questions present in each exercise. Experts, while creating the solutions, have followed the latest syllabus and framed it in accordance with the exam pattern of the CBSE Board. Students can download the readily available PDF of RD Sharma Solutions from the links provided below and start practising offline. Now, let us have a look at the concepts discussed in this chapter.

• Random experiments
• Sample spaces
• Event
• Algebra of events
• Types of events
• Axiomatic approach to probability
• Addition theorems on probability

RD Sharma Solutions for Class 11 Maths Chapter 33 – Probability

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EXERCISE 33.1 PAGE NO: 33.6

1. A coin is tossed once. Write its sample space.

Solution:

Given: A coin is tossed once.

We know that the coin is tossed only once.

Then, there are two probabilities either Head (H) or Tail (T).

So,

S = {H, T}

∴ The sample space is {H, T}

2. If a coin is tossed two times, describe the sample space associated with this experiment.

Solution:

Given: If the coin is tossed twice times.

We know that two coins are tossed, which means two probabilities will occur at the same time.

So,

S = {HT, TH, HH, TT}

∴ Sample space is {HT, HH, TT, TH}

3. If a coin is tossed three times (or three coins are tossed together), then describe the sample space for this experiment.

Solution:

Given: If a coin is tossed three times.

We know that the coins are tossed three times, and then the no. of samples is

2³ = 8

So,

S = {HHH, TTT, HHT, HTH, THH, HTT, THT, TTH}

∴ The sample space is {HHH, TTT, HHT, HTH, THH, HTT, THT, TTH}

4. Write the sample space for the experiment of tossing a coin four times.

Solution:

Given: A coin is tossed four times.

We know that the coins are tossed four times, then the no. of samples

2⁴ = 16

So,

S = {HHHH, TTTT, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH}

∴ The sample space is {HHHH, TTTT, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH}

5. Two dice are thrown. Describe the sample space of this experiment.

Solution:

Given: Two dice are thrown.

We know there are 6 faces on a die. Which contains (1, 2, 3, 4, 5, 6).

Here two dice are thrown, and then we have two faces of dice (one of each).

So, the total sample space will be 6² = 36

∴ The sample space is

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

6. What is the total number of elementary events associated with the random experiment of throwing three dice together?

Solution:

Given: Three dice are rolled together.

We know that three dice are thrown together. And there are 6 faces on a die.

So, the total numbers of the elementary event on throwing three dice are

6 × 6 × 6 = 216

∴ The total number of elementary events is 216.

7. A coin is tossed, and then a die is thrown. Describe the sample space for this experiment.

Solution:

Given: A coin is tossed, and a die is thrown.

We know that the coin is tossed, and the die is thrown.

So, when the coin is tossed there will be 2 events, either Head or Tail.

And, when the die is thrown, then there will be 6 faces (1, 2, 3, 4, 5, 6)

Then,

The total number of sample spaces together is 2 × 6 = 12

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

∴ The sample space is {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

8. A coin is tossed, and then a die is rolled only in case a head is shown on the coin. Describe the sample space for this experiment.

Solution:

Given: A coin is tossed, and the die is rolled.

We know that we have a coin and a die.

So, when a coin is tossed, there will be 2 events – head and tail.

According to the question, if the head occurs on the coin, then the die will be rolled out otherwise not.

So, the sample spaces are

S = {(T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

∴ The sample space is {T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)1}

9. A coin is tossed twice. If the second throw results in a tail, a die is thrown. Describe the sample space for this experiment.

Solution:

Given: A coin is tossed twice. If the second throw results in a tail, a die is thrown.

When a coin is tossed twice, then sample spaces for the only coin will be: {HH, TT, HT, TH}

Now, according to the question, when we get tail in the second throw, then a dice is thrown.

So, the total number of elementary events is 2 + (2×6) = 14

And sample space will be

S = {HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3), (TT, 4), (TT, 5), (TT, 6)}

∴ The sample space is {HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3), (TT, 4), (TT, 5), (TT, 6)}

10. An experiment consists of tossing a coin and then tossing it the second time if the head occurs. If a tail occurs on the first toss, then a die is tossed once. Find the sample space.

Solution:

Given: A coin is tossed, and a die is rolled.

In the given experiment, the coin is tossed, and if the outcome is tail, then the die will be rolled.

The possible outcome for coin is 2 = {H, T}

And, the possible outcome for die is 6 = {1, 2, 3, 4, 5, 6}

If the outcome for the coin is tail, then sample space is S1= {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)}

If the outcome is head then the sample space is S2 = {(H, H) (H, T)}

So, the required outcome sample space is S = S1 ⋃ S2

S = {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (H, H) (H, T)}

∴ The sample space for the given experiment is {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (H, H) (H, T)}

11. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red 3 black balls; if it shows a head, we throw a die. Find the sample space of this experiment.

Solution:

Given: A coin is tossed, and there is a box which contains 2 red and 3 black balls.

When the coin is tossed, there are 2 outcomes {H, T}

According to the question, if the tail is turned up, the ball is drawn from a box.

So, sample for this experiment S₁ = {(T, R₁) (T, R₂) (T, B₁) (T, B₂) (T, B₃)}

Now, if the head is turned up and then the die is rolled.

So, sample space for this experiment S₂ = {(H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)}

The required sample space will be S = S₁ ⋃ S₂

So,

S = {(T, R₁), (T, R₂), (T, B₁), (T, B₂), (T, B₃), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

∴ S is the elementary events associated with the given experiment.

12. A coin is tossed repeatedly until a tail comes up for the first time. Write the sample space for this experiment.

Solution:

Given: A coin is tossed repeatedly until it comes up for the first time.

In the given Experiment, a coin is tossed, and if the outcome is tail, the experiment is over.

And, if the outcome is head, then the coin is tossed again.

In the second toss also, if the outcome is tail, then the experiment is over; otherwise, the coin is tossed again.

This process continues indefinitely.

So, the sample space for this experiment is

S = {T, HT, HHT, HHHT, HHHHT…}

∴ The sample space for the given experiment is {T, HT, HHT, HHHT, HHHHT…}

EXERCISE 33.2 PAGE NO: 33.15

1. A coin is tossed. Find the total number of elementary events and also the total number of events associated with the random experiment.

Solution:

Given: A coin is tossed.

When a coin is tossed, there will be two possible outcomes, Head (H) and Tail (T).

Since the no. of elementary events is 2 {H, T}

We know if there are n elements in a set, then the number of total elements in its subset is 2n.

So, the total number of the experiment is 4,

There are 4 subset of S = {H}, {T}, {H, T} and Փ

∴ There is a total of 4 events in a given experiment.

2. List all events associated with the random experiment of tossing two coins. How many of them are elementary events?

Solution:

Given: Two coins are tossed once.

We know when two coins are tossed then, the no. of possible outcomes is 2² = 4

So, the sample spaces are {HH, HT, TT, TH}

∴ There is a total of 4 events associated with the given experiment.

3. Three coins are tossed once. Describe the following events associated with this random experiment:
A = Getting three heads, B = Getting two heads and one tail, C = Getting three tails, D = Getting a head on the first coin.
(i) Which pairs of events are mutually exclusive?
(ii) Which events are elementary events?
(iii) Which events are compound events?

Solution:

Given: There are three coins tossed once.

When three coins are tossed, then the sample spaces are

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

So, according to the question,

A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

Now, A⋂ B = Փ,

A ⋂ C = Փ,

A ⋂ D = {HHH}

B⋂ C = Փ,

B ⋂ D = {HHT, HTH}

C ⋂ D = Փ

We know that if the intersection of two sets is null or empty, it means both sets are Mutually Exclusive.

(i) Events A and B, Events A and C, Events B and C and events C and D are mutually exclusive.

(ii) Here, we know if an event has only one sample point of a sample space, then it is called an elementary event.

So, A and C are elementary events.

(iii) If there is an event that has more than one sample point of a sample space, it is called a compound event.

Since, B ⋂ D = {HHT, HTH}

So, B and D are compound events.

4. In a single throw of a die, describe the following events:
(i) A = Getting a number less than 7
(ii) B = Getting a number greater than 7
(iii) C = Getting a multiple of 3
(iv) D = Getting a number less than 4
(v) E = Getting an even number greater than 4.
(vi) F = Getting a number not less than 3.
Also, find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and.

Solution:

Given: A dice is thrown once.

Let us find the given events and also find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and
.

S = {1, 2, 3, 4, 5, 6}

According to the subparts of the question, we have certain events

(i) A = getting a number below 7

So, the sample spaces for A are

A = {1, 2, 3, 4, 5, 6}

(ii) B = Getting a number greater than 7

So, the sample spaces for B are:

B = {Փ}

(iii) C = Getting multiple of 3

So, the Sample space of C is

C = {3, 6}

(iv) D = Getting a number less than 4

So, the sample space for D is

D = {1, 2, 3}

(v) E = Getting an even number greater than 4.

So, the sample space for E is

E = {6}

(vi) F = Getting a number not less than 3.

So, the sample space for F is

F = {3, 4, 5, 6}

Now,

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

A ⋃ B = {1, 2, 3, 4, 5, 6}

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

A ⋂ B = {Փ}

B = {Փ} and C = {3, 6}

B ⋂ C = {Փ}

F = {3, 4, 5, 6} and E = {6}

E ⋂ F = {6}

E = {6} and D = {1, 2, 3}

D ⋂ F = {3}

5. Three coins are tossed. Describe
(i) two events, A and B, which are mutually exclusive.
(ii) three events, A, B and C, which are mutually exclusive and exhaustive.
(iii) two events, A and B, which are not mutually exclusive.
(iv) two events, A and B, which are mutually exclusive but not exhaustive.

Solution:

Given: Three coins are tossed.

When three coins are tossed, then the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Now, the subparts are

(i) The two events which are mutually exclusive are when,

A: getting no tails

B: getting no heads

Then, A = {HHH} and B = {TTT}

So, the intersection of this set will be null. Or, the sets are disjoint.

(ii) Three events which are mutually exclusive and exhaustive are

A: getting no heads

B: getting exactly one head

C: getting at least two head

So, A = {TTT} B = {TTH, THT, HTT} and C = {HHH, HHT, HTH, THH}

Since, A ⋃ B = B ⋂ C = C ⋂ A = Փ and

A⋃ B⋃ C = S

(iii) The two events that are not mutually exclusive are

A: getting three heads

B: getting at least 2 heads

So, A = {HHH} B = {HHH, HHT, HTH, THH}

Hence, A ⋂ B = {HHH} = Փ

(iv) The two events, which are mutually exclusive but not exhaustive, are

A: getting exactly one head

B: getting exactly one tail

So, A = {HTT, THT, TTH} and B = {HHT, HTH, THH}

It is because A ⋂ B = Փ but A⋃ B ≠ S

6. A die is thrown twice. Each time the number appearing on it is recorded. Describe the following events:
(i) A = Both numbers are odd.
(ii) B = Both numbers are even
(iii) C = sum of the numbers is less than 6.
Also, find A ∪ B, A ∩ B, A ∪ C, A ∩ C. Which pairs of events are mutually exclusive?

Solution:

Given: A dice is thrown twice. And each time number appearing on it is recorded.

When the dice are thrown twice, then the number of sample spaces is 6² = 36

Now,

The possibility both odd numbers are:

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}

Since the possibility of both even numbers is:

B = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

And the possible outcome of the sum of the numbers is less than 6.

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

Hence,

(AՍB) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)}

(AՌB) = {Փ}

(AUC) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (1, 2)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

(AՌC) = {(1, 1), (1, 3), (3, 1)}

∴ (AՌB) = Փ and (AՌC) ≠ Փ, A and B are mutually exclusive, but A and C are not.

EXERCISE 33.3 PAGE NO: 33.45

1. Which of the following cannot be a valid assignment of probability for elementary events or outcomes of sample space S = {w₁, w₂, w₃, w₄, w₅, w₆, w₇}:

Solution:

For each event to be a valid assignment of probability.

The probability of each event in the sample space should be less than 1, and the sum of the probability of all the events should be exactly equal to 1.

(i) It is valid as each P (w_i) (for i=1 to 7) lies between 0 to 1, and the sum of P (w₁) =1

(ii) It is valid as each P (w_i) (for i=1 to 7) lies between 0 to 1, and the sum of P (w₁) =1

(iii) It is not valid as the sum of P (w_i) = 2.8, which is greater than 1.

(iv) it is not valid as P (w₇) = 15/14, which is greater than 1.

2. A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3

Solution:

Given: A die is thrown.

The total number of outcomes is six, n (S) = 6

By using the formula,

P (E) = favourable outcomes / total possible outcomes

(i) Let E be the event of getting a prime number.

E = {2, 3, 5}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 6

= ½

(ii) Let E be the event of getting 2 or 4.

E = {2, 4}

n (E) = 2

P (E) = n (E) / n (S)

= 2 / 6

= 1/3

(iii) Let E be the event of getting a multiple of 2 or 3.

E = {2, 3, 4, 6}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 6

= 2/3

3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) an even number on first
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11, as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) neither a doublet nor a total of 10
(xiii) odd number on the first and 6 on the second
(xiv) a number greater than 4 on each die
(xv) a total of 9 or 11
(xvi) a total greater than 8

Solution:

Given: A pair of dice has been thrown, so the number of elementary events in sample space is 6² = 36

n (S) = 36

By using the formula,

P (E) = favourable outcomes / total possible outcomes

(i) Let E be the event that the sum 8 appears.

E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

n (E) = 5

P (E) = n (E) / n (S)

= 5 / 36

(ii) Let E be the event of getting a doublet.

E = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(iii) Let E be the event of getting a doublet of prime numbers.

E = {((2, 2) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(iv) Let E be the event of getting a doublet of odd numbers.

E = {(1, 1) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(v) Let E be the event of getting a sum greater than 9.

E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(vi) Let E be the event of getting even on the first die.

E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

n (E) = 18

P (E) = n (E) / n (S)

= 18 / 36

= ½

(vii) Let E be the event of getting even on one and multiple of three on the other.

E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}

n (E) = 11

P (E) = n (E) / n (S)

= 11 / 36

(viii) Let E be the event of getting neither 9 nor 11 as the sum.

E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(ix) Let E be the event of getting a sum less than 6.

E = {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}

n (E) = 10

P (E) = n (E) / n (S)

= 10 / 36

= 5/18

(x) Let E be the event of getting a sum less than 7.

E = {(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1)}

n (E) = 15

P (E) = n (E) / n (S)

= 15 / 36

= 5/12

(xi) Let E be the event of getting more than 7.

E = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}

n (E) = 15

P (E) = n (E) / n (S)

= 15 / 36

= 5/12

(xii) Let E be the event of getting neither a doublet nor a total of 10.

E′ be the event that either a double or a sum of ten appears.

E′ = {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}

n (E′) = 8

P (E′) = n (E′) / n (S)

= 8 / 36

= 2/9

So, P (E) = 1 – P (E′)

= 1 – 2/9

= 7/9

(xiii) Let E be the event of getting an odd number on the first and 6 on the second.

E = {(1,6) (5,6) (3,6)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(xiv) Let E be the event of getting greater than 4 on each die.

E = {(5,5) (5,6) (6,5) (6,6)}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 36

= 1/9

(xv) Let E be the event of getting a total of 9 or 11.

E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(xvi) Let E be the event of getting a total greater than 8.

E = {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}

n (E) = 10

P (E) = n (E) / n (S)

= 10 / 36

= 5/18

4. In a single throw of three dice, find the probability of getting a total of 17 or 18

Solution:

Given: The dice are thrown.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Total number of possible outcomes is 6³=216

So, n (S) = 216

Let E be the event of getting a total of 17 or 18.

E = {(6, 6, 5) (6, 5, 6) (5, 6, 6) (6, 6, 6)}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 216

= 1/54

5. Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail

Solution:

Given: Three coins are tossed together.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Total number of possible outcomes is 2³ = 8

(i) Let E be the event of getting exactly two heads.

E = {(H, H, T) (H, T, H) (T, H, H)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 8

(ii) Let E be the event of getting at least two heads.

E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}

n (E)=4

P (E) = n (E) / n (S)

= 4 / 8

= ½

(iii) Let E be the event of getting at least one head and one tail.

E = {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 8

= ¾

6. What is the probability that an ordinary year has 53 Sundays?

Solution:

Given: A year which includes 52 weeks and one day.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of that one day being Sunday.

Total number of possible outcomes is 7.

n(S) = 7

E = {M, T, W, T, F, S, SU}

n (E) = 1

P (E) = n (E) / n (S)

= 1 / 7

7. What is the probability that a leap year has 53 Sundays and 53 Mondays?

Solution:

Given: A leap year which includes 52 weeks and two days.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of the remaining two days being Sunday and Monday.

S = {MT, TW, WT, TF, FS, SSu, SuM}

n (S) = 7

E= {SuM}

n (E) = 1

P (E) = n (E) / n (S)

= 1 / 7

8. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(i) All three balls are white
(ii) All three balls are red
(iii) One ball is red, and two balls are white

Solution:

Given: A bag contains 8 red and 5 white balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Total number of ways of drawing three balls at random is ¹³C₃

n (S) = 286

(i) Let E be the event of getting all white balls.

E= {(W) (W) (W)}

n (E)= ⁵C₃=10

P (E) = n (E) / n (S)

= 10 / 286

= 5/143

(ii) Let E be the event of getting all red balls.

E = {(R) (R) (R)}

n (E)= ⁸C₃ =56

P (E) = n (E) / n (S)

= 56 / 286

= 28/143

(iii) Let E be the event of getting one red and two white balls.

E = {(R…. 80^th R)}

n (E)= ⁸C₁⁵C₂ = 80

P (E) = n (E) / n (S)

= 80 / 286

= 40/143

9. In a single throw of three dice, find the probability of getting the same number on all the three dice

Solution:

Given: Three dice are rolled over.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of getting the same number on all three dice.

Total number of possible outcomes is 6³=216

n (S) = 216

Let E be the event of getting the same number on all three dice.

E = {(1,1,1) (2,2,2) (3,3,3) (4,4,4) (5,5,5) (6,6,6)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 216

= 1/36

10. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10

Solution:

Given: Two unbiased dice are thrown.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

So, we now have to determine the probability of getting the sum of digits on dice greater than 10.

Total number of possible outcomes is 6²=36

n (S) = 36

Let E be the event of getting the same number on all three dice.

E = {(5,6) (6,5) (6,6)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

11. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:

(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither an ace nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) a diamond card
(ix) not a diamond card
(x) a black card
(xi) not an ace
(xii) not a black card

Solution:

Given: Pack of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We know that a card is drawn from a pack of 52 cards, so the number of elementary events in the sample space is

n (S) = ⁵²C₁ = 52

(i) Let E be the event of drawing a black king

n (E) =²C₁ =2 (there are two black kings, one of a spade and the other of a club)

P (E) = n (E) / n (S)

= 2 / 52

= 1/26

(ii) Let E be the event of drawing a black card or a king

n (E) = ²⁶C₁+⁴C₁–²C₁= 28

P (E) = n (E) / n (S)

= 28 / 52

= 7/13

(iii) Let E be the event of drawing a black card and a king.

n (E) =²C₁ = 2 (there are two black kings, one of spade and the other of a club)

P (E) = n (E) / n (S)

= 2 / 52

= 1/26

(iv) Let E be the event of drawing a jack, queen or king.

n (E) = ⁴C₁+⁴C₁+⁴C₁ = 12

P (E) = n (E) / n (S)

= 12 / 52

= 3/13

(v) Let E be the event of drawing neither a heart nor a king.

Now let us consider E′ as the event in either a heart or king appears.

n (E′) = ⁶C₁+⁴C₁-1=16 (there is a heart king, so it is deducted.)

P (E′) = n (E′) / n (S)

= 16 / 52

= 4/13

So, P (E) = 1 – P (E′)

= 1 – 4/13

= 9/13

(vi) Let E be the event of drawing a spade or king.

n (E)=¹³C₁+⁴C₁-1=16

P (E) = n (E) / n (S)

= 16 / 52

= 4/13

(vii) Let E be the event of drawing neither an ace nor a king.

Now, let us consider E′ as the event in which either an ace or king appears.

n(E′) = ⁴C₁+⁴C₁ = 8

P (E′) = n (E′) / n (S)

= 8 / 52

= 2/13

So, P (E) = 1 – P (E′)

= 1 – 2/13

= 11/13

(viii) Let E be the event of drawing a diamond card.

n (E)=¹³C₁=13

P (E) = n (E) / n (S)

= 13 / 52

= ¼

(ix) Let E be the event of drawing, not a diamond card.

Now, let us consider E′ as the event that a diamond card appears.

n (E′) =¹³C₁=13

P (E′) = n (E′) / n (S)

= 13 / 52

= 1/4

So, P (E) = 1 – P (E′)

= 1 – 1/4

= ¾

(x) Let E be the event of drawing a black card.

n (E) =²⁶C₁ = 26 (spades and clubs)

P (E) = n (E) / n (S)

= 26 / 52

= ½

(xi) Let E be the event of drawing, not an ace.

Now, let us consider E′ as the event that an ace card appears.

n (E′) = ⁴C₁ = 4

P (E′) = n (E′) / n (S)

= 4 / 52

= 1/13

So, P (E) = 1 – P (E′)

= 1 – 1/13

=12/13

(xii) Let E be the event of not drawing a black card.

n (E) = ²⁶C₁ = 26 (red cards of hearts and diamonds)

P (E) = n (E) / n (S)

= 26 / 52

= ½

12. In shutting a pack of 52 playing cards, four are accidentally dropped; find the chance that the missing cards should be one from each suit

Solution:

Given: A pack of 52 cards from which 4 are dropped.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We now have to find the probability that the missing cards should be one from each suit.

We know that, from a well-shuffled pack of cards, 4 cards missed out on total possible outcomes are

n (S) = ⁵²C₄ = 270725

Let E be the event that four missing cards are from each suit.

n (E) = ¹³C₁×¹³C₁×¹³C₁×¹³C₁ = 13⁴

P (E) = n (E) / n (S)

= 13⁴ / 270725

= 2197/20825

13. From a deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honours of the same suit

Solution:

Given: A deck of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

We now have to find the probability that all the face cards of the same suits are drawn.

Total possible outcomes are

n (S) = ⁵²C₄

Let E be the event that all the cards drawn are face cards of the same suit.

n (E)=4×⁴C₄=4

P (E) = n (E) / n (S)

= 4 / 270725

14. Tickets numbered from 1 to 20 are mixed up together, and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?

Solution:

Given: Numbered tickets from 1 to 20.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

To find the probability of the ticket drawn having a number which is a multiple of 3 or 7.

We know that one ticket is drawn from a lot of mixed numbers.

Total possible outcomes are

n (S) = ²⁰C₁ = 20

Let E be the event of getting a ticket which has a number that is multiple of 3 or 7.

E = {3,6,9,12,15,18,7,14}

n (E) = 8

P (E) = n (E) / n (S)

= 8 / 20

= 2/5

15. A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white, and one is blue.

Solution:

Given: A bag containing 6 red, 4 white and 8 blue balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Three balls are drawn, so we have to find the probability that one is red, one is white, and one is blue.

Total number of outcomes for drawing 3 balls is ¹⁸C₃

n (S) = ¹⁸C₃ = 816

Let E be the event that one red, one white and one blue ball is drawn.

n (E) = ⁶C₁⁴C₁⁸C₁ = 192

P (E) = n (E) / n (S)

= 192 / 816

= 4/17

16. A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:
(i) both the balls are white
(ii) one ball is black, and the other red
(iii) both the balls are of the same colour

Solution:

Given: A bag containing 7 white, 5 black and 4 red balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random, therefore

Total possible outcomes are ¹⁶C₂

n (S) = 120

(i) Let E be the event of getting both white balls.

E = {(W) (W)}

n (E) = ⁷C₂ = 21

P (E) = n (E) / n (S)

= 21 / 120

= 7/40

(ii) Let E be the event of getting one black and one red ball.

E = {(B) (R)}

n (E) = ⁵C₁⁴C₁ = 20

P (E) = n (E) / n (S)

= 20 / 120

= 1/6

(iii) Let E be the event of getting both balls of the same colour.

E = {(B) (B)} or {(W) (W)} or {(R) (R)}

n (E) = ⁷C₂+⁵C₂+⁴C₂ = 37

P (E) = n (E) / n (S)

= 37 / 120

17. A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) one is red, and two are white
(ii) two are blue, and one is red
(iii) one is red

Solution:

Given: A bag containing 6 red, 4 white and 8 blue balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random.

Total possible outcomes are ¹⁸C₃

n (S) = 816

(i) Let E be the event of getting one red and two white balls.

E = {(W) (W) (R)}

n (E) = ⁶C₁⁴C₂ = 36

P (E) = n (E) / n (S)

= 36 / 816

= 3/68

(ii) Let E be the event of getting two blue and one red.

E = {(B) (B) (R)}

n (E) = ⁸C₂⁶C₁ = 168

P (E) = n (E) / n (S)

= 168 / 816

= 7/34

(iii) Let E be the event that one of the balls must be red.

E = {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}

n (E) = ⁶C₁⁴C₁⁸C₁+⁶C₁⁴C₂+⁶C₁⁸C₂ = 396

P (E) = n (E) / n (S)

= 396 / 816

= 33/68

18. Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:
(i) just one ace
(ii) at least one ace?

Solution:

Given: Five cards are drawn from a pack of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Five cards are drawn at random.

Total possible outcomes are ⁵²C₅

n (S) = 2598960

(i) Let E be the event that exactly only one ace is present.

n (E) = ⁴C₁⁴⁸C₄ = 778320

P (E) = n (E) / n (S)

= 778320 / 2598960

= 3243/10829

(ii) Let E be the event that at least one ace is present.

E = {1 or 2 or 3 or 4 ace(s)}

n (E) = ⁴C₁⁴⁸C₄+⁴C₂⁴⁸C₃+⁴C₃⁴⁸C₂+⁴C₄⁴⁸C₁ = 886656

P (E) = n (E) / n (S)

= 886656 / 2598960

= 18472/54145

19. The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. What is the probability that they belong to different suits?

Solution:

Given: The face cards are removed from a full pack of 52.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Four cards are drawn from the remaining 40 cards, so we have to find the probability that all of them belong to a different suit.

Total possible outcomes of drawing four cards are ⁴⁰C₄

n (S) =  ⁴⁰C₄ = 91390

Let E be the event that 4 cards belong to a different suit.

n (E) = ¹⁰C₁¹⁰C₁¹⁰C₁¹⁰C₁ = 10000

P (E) = n (E) / n (S)

= 10000 / 91390

= 1000/9139

20. There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is it that it is a woman?

Solution:

Given: There are four men and six women on the city councils.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

From the city council, one person is selected as a council member, so we have to find the probability that it is a woman.

Total possible outcomes of selecting a person are ¹⁰C₁

n (S)= ¹⁰C₁ = 10

Let E be the event that it is a woman.

n (E) = ⁶C₁ = 6

P (E) = n (E) / n (S)

= 6 / 10

= 3/5

21. A box contains 100bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:
(i) all 10 are defective
(ii) all 10 are good
(iii) at least one is defective
(iv) none is defective

Solution:

Given: A box contains 100 bulbs, 20 of which are defective.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Ten bulbs are drawn at random for inspection.

Total possible outcomes are ¹⁰⁰C₁₀

n (S) = ¹⁰⁰C₁₀

(i) Let E be the event that all ten bulbs are defective.

n (E) = ²⁰C₁₀

P (E) = n (E) / n (S)

= ²⁰C₁₀ / ¹⁰⁰C₁₀

(ii) Let E be the event that all ten good bulbs are selected.

n (E) = ⁸⁰C₁₀

P (E) = n (E) / n (S)

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

(iii) Let E be the event that at least one bulb is defective.

E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs

Let E′ be the event that none of the bulbs is defective.

n (E′) = ⁸⁰C₁₀

P (E′) = n (E′) / n (S)

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

So, P (E) = 1 – P (E′)

= 1 – ⁸⁰C₁₀ / ¹⁰⁰C₁₀

(iv) Let E be the event that none of the selected bulbs is defective.

n (E) = ⁸⁰C₁₀

P (E) = n (E) / n (S)

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

22. Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together

Solution:

Given: The word ‘SOCIAL’.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

In the random arrangement of the alphabet of the word “SOCIAL”, we have to find the probability that vowels come together.

Total possible outcomes of arranging the alphabet are 6!

n (S) = 6!

Let E be the event that vowels come together.

The vowels in SOCIAL are A, I, O.

So, the number of ways to arrange them where (A, I, O) come together.

n (E) = 4! × 3!

P (E) = n (E) / n (S)

= [4! × 3!] / 6!

= 1/5

23. The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?

Solution:

Given: The word ‘CLIFTON’.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

In the random arrangement of the alphabet of the word “CLIFTON”, we have to find the probability that vowels come together.

Total possible outcomes of arranging the alphabet are 7!

n (S) =7!

Let E be the event that vowels come together.

The vowels in CLIFTON are I, O.

Number of ways to arrange them where (I, O) come together.

n (E)= 6! × 2!

P (E) = n (E) / n (S)

= [6! × 2!] / 7!

= 2/7

EXERCISE 33.4 PAGE NO: 33.67

1. (a) If A and B be mutually exclusive events associated with a random experiment such that P (A) = 0.4 and P (B) = 0.5, then find:
(i) P(A ∪ B)

Solution:

Given: A and B are two mutually exclusive events.

P (A) = 0.4 and P (B) = 0.5

By definition of mutually exclusive events, we know that

P (A ∪ B) = P (A) + P (B)

Now, we have to find

(i) P (A ∪ B) = P (A) + P (B) = 0.5 + 0.4 = 0.9

(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.9

= 0.1

(iii) P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]

P (only B) = P (B) – P (A ∩ B)

As A and B are mutually exclusive, so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A′ ∩ B) = P (B) = 0.5

(iv) P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence, this indicates only A.]

P (only A) = P (A) – P (A ∩ B)

As A and B are mutually exclusive, so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A ∩ B′) = P (A) = 0.4

(b) A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35. Find (i) P (A ∪ B)

Solution:

Given: A and B are two events.

P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35

By definition of P (A or B) under the axiomatic approach, we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

Now, we have to find

(i) P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 0.54 + 0.69 – 0.35

= 0.88

(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.88

= 0.12

(iii) P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence, this indicates only A.]

P (only A) = P (A) – P (A ∩ B)

∴ P (A ∩ B′) = P (A) – P (A ∩ B)

= 0.54 – 0.35

= 0.19

(iv) P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence, this indicates only B.]

P (only B) = P (B) – P (A ∩ B)

∴ P (A′ ∩ B) = P (B) – P (A ∩ B)

= 0.69 – 0.35

= 0.34

(c) Fill in the blanks in the following table:

Solution:

(i) By definition of P (A or B) under the axiomatic approach, we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

By using data from the table, we get

∴ P (A ∪ B) = 1/3 + 1/5 – 1/15

= 8/15 – 1/15

= 7/15

(ii) By definition of P (A or B) under the axiomatic approach, we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (B) = P (A ∪ B) + P (A ∩ B) – P (A)

By using data from the table, we get

∴ P (B) = 0.6 + 0.25 – 0.35

= 0.5

(iii) By definition of P (A or B) under the axiomatic approach, we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∩ B) = P (B) + P (A) – P (A ∪ B)

By using data from a table, we get

∴ P (A ∩ B) = 0.5 + 0.35 – 0.7

= 0.15

Hence the table is:

2. If A and B are two events associated with a random experiment such that P (A) = 0.3, P (B) = 0.4 and P (A ∪ B) = 0.5, find P (A ∩ B).

Solution:

Given: A and B are two events.

P (A) = 0.3, P (B) = 0.5 and P (A ∪ B) = 0.5

Now we need to find P (A ∩ B).

By definition of P (A or B) under the axiomatic approach (also called addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (A ∩ B) = P (A) + P (B) – P (A ∪ B)

P (A ∩ B) = 0.3 + 0.4 – 0.5

= 0.7 – 0.5

= 0.2

∴ P (A ∩ B) is 0.2

3. If A and B are two events associated with a random experiment such that P (A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2, find P (A ∪ B).

Solution:

Given: A and B are two events.

P (A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2

Now we need to find P (A ∪ B).

By definition of P (A or B) under the axiomatic approach (also called addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∪ B) = 0.5 + 0.3 – 0.2

= 0.8 – 0.2

= 0.6

∴ P (A ∪ B) is 0.6

Solution:

Given: A and B are two events.

P (A′) = 0.5, P (A ∩ B) = 0.3 and P (A ∪ B) = 0.8

Since, P (A′) = 1 – P (A)

P (A) = 1 – 0.5

= 0.5

Now we need to find P (B).

By definition of P (A or B) under the axiomatic approach (also called addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (B) = P (A ∪ B) + P (A ∩ B) – P (A)

P (B) = 0.8 + 0.3 – 0.5

= 1.1 – 0.5

= 0.6

∴ P (B) is 0.6

5. Given two mutually exclusive events, A and B, such that P (A) = 1/2 and P (B) = 1/3, find P (A or B).

Solution:

Given: A and B are two mutually exclusive events.

P (A) = 1/2 and P (B) = 1/3

Now we need to find P (A ‘or’ B).

P (A or B) = P (A ∪ B)

So by definition of mutually exclusive events, we know that

P (A ∪ B) = P (A) + P (B)

= 1/2 + 1/3

= 5/6

∴ P (A ∪ B) is 5/6

6. There are three events, A, B and C, one of which must and only one can happen, the odds are 8 to 3 against A, 5 to 2 against B, and find the odds against C.

Solution:

As out of 3 events, A, B and C, only one can happen at a time, which means no event have anything in common.

∴ We can say that A, B and C are mutually exclusive events.

So, by definition of mutually exclusive events, we know that

P (A ∪ B ∪ C) = P (A) + P (B) + P (C)

According to the question, one event must happen.

So, A or B or C is a sure event.

∴ P (A ∪ B ∪ C) = 1 … Equation (1)

We need to find odd against C.

Given:

Odd against A = 8/3

8 P (A) = 3 – 3 P (A)

11 P (A) = 3

∴ P (A) = 3/11 … Equation (2)

Similarly, we are given: Odd against B = 5/2

5 P (B) = 2 – 2 P(B)

7 P (B) = 2

∴ P (B) =2/7 …Equation (3)

From equations 1, 2 and 3, we get

P (C) = 1 – 3/11 – 2/7

= [77-21-22]/77

= 34/77

So, P (C′) = 1 – (34/77)

= 43/77

∴ Odd against C:

7. One of the two events must happen. Given that the chance of one is two-thirds of the other, find the odds in favour of the other.

Solution:

Let A and B be two events.

As out of 2 events, A and B, only one can happen at a time, which means no event have anything in common.

∴ We can say that A and B are mutually exclusive events.

So, by definition of mutually exclusive events, we know that

P (A ∪ B) = P (A) + P (B)

According to the question, one event must happen.

A or B is a sure event.

So, P (A ∪ B) = P (A) + P (B) = 1 … Equation (1)

Given: P (A) = (2/3) P (B)

We have to find the odds in favour of B.

P (B) = 3/5

So, P (B′) = 1 – 3/5

= 2/5

∴ Odd in favour of B:

8. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of its being a spade or a king.

Solution:

Given: As a card is drawn from a deck of 52 cards.

Let ‘S’ denotes the event of the card being a spade and ‘K’ denote the event of the card is a King.

As we know that a deck of 52 cards contains 4 suits (Heart, Diamond, Spade and Club), each having 13 cards. The deck has 4 king cards, one from each suit.

We know that the probability of an event E is given as-

By using the formula,

P (E) = favourable outcomes / total possible outcomes

= n (E) / n (S)

Where n (E) = numbers of elements in event set E

And n (S) = numbers of elements in sample space.

Hence,

P (S) = n (spade) / total number of cards

= 13 / 52

= ¼

P (K) = 4/52

= 1/13

And P (S ⋂ K) = 1/52

We need to find the probability that the card is a spade or king, i.e.,

P (Spade ‘or’ King) = P(S ∪ K)

So, by definition of P (A or B) under the axiomatic approach (also called addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (S ∪ K) = P (S) + P (K) – P (S ∩ K)

= ¼ + 1/13 – 1/52

= 17/52 – 1/52

= 16/52

= 4/13

∴ P (S ∪ K) = 4/13

9. In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

Solution:

In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n (S) = 36

Where ‘S’ represents sample space.

Let ‘A’ denotes the event of getting a double.

So, A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

P (A) = n (A) / n (S)

= 6/36

= 1/6

And ‘B’ denotes the event of getting a total of 9.

So, B = {(3,6), (6,3), (4,5), (5,4)}

P (B) = n (B) / n (S)

= 4/36

= 1/9

We need to find the probability of neither the event of getting neither a doublet nor a total of 9.

P (A′ ∩ B′) =?

As, P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s theorem}

P (A′ ∩ B′) = 1 – P (A ∪ B)

By using the definition of P (E or F) under the axiomatic approach (also called the addition theorem), we know that

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

∴ P (A ∪ B) = 1/6 + 1/9 + 0

= 5/18 {Since, P (A ∩ B) = 0 since nothing is common in sets A and B.

So, n (A ∩ B) = 0}

Hence,

P (A′ ∩ B′) = 1 – (5/18)

= 13/18

10. A natural number is chosen at random from amongst the first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Solution:

Given: Sample space is the set of the first 500 natural numbers.

n (S) = 500

Let ‘A’ be the event of choosing the number such that it is divisible by 3.

n (A) = [500/3]

= [166.67]

= 166 {where [.] represents Greatest integer function}

P (A) = n (A) / n (S)

= 166/500

= 83/250

Let ‘B’ be the event of choosing the number such that it is divisible by 5.

n (B) = [500/5]

= [100]

= 100 {where [.] represents Greatest integer function}

P (B) = n (B) / n (S)

= 100/500

= 1/5

Now, we need to find the P (such that number chosen is divisible by 3 or 5)

P (A or B) = P (A ∪ B)

By using the definition of P (E or F) under the axiomatic approach (also called the addition theorem), we know that

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

n(A ∩ B) = [500/15]

= [33.34]

= 33

P (A ∩ B) = n(A ∩ B) / n (S)

= 33/500

∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 83/250 + 1/5 – 33/500

= [166 + 100 – 33]/500

= 233/500

11. A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3?

Solution:

If a dice is thrown twice, it has a total of (6 × 6) = 36 possible outcomes.

If S represents the sample space, then

n (S) = 36

Let ‘A’ represent events the event such that 3 comes in the first throw.

A = {(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)}

P (A) = n (A) / n (S)

= 6 /36

= 1/6

Let ‘B’ represent events the event such that 3 comes in the second throw.

B = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}

P (B) = n (B) / n (S)

= 6 /36

= 1/6

It is clear that (3,3) is common in both events, so

P (A ∩ B) = n (A ∩ B) / n (S)

= 1/ 36

Now we need to find the probability of the event, such that at least one of the 2 throws gives 3, i.e., P (A or B) = P (A ∪ B)

By using the definition of P (E or F) under the axiomatic approach (also called the addition theorem), we know that

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

So, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 1/6 + 1/6 – 1/36

= 1/3 – 1/36

= 11/36

∴ P (at least one of the two throws comes to be 3) is 11/36

12. A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

Solution:

Given: As a card is drawn from a deck of 52 cards.

Let ‘S’ denotes the event of a card being a spade and ‘K’ denote the event of a card is an ace.

As we know that a deck of 52 cards contains 4 suits (Heart, Diamond, Spade and Club), each having 13 cards. The deck has 4 ace cards, one from each suit.

We know that the probability of an event E is given as

By using the formula,

P (E) = favourable outcomes / total possible outcomes

= n (E) / n (S)

Where n (E) = numbers of elements in event set E

And n (S) = numbers of elements in sample space

Hence,

P (S) = n (spade) / total number of cards

= 13 / 52

= ¼

P (K) = 4/52

= 1/13

And P (S ⋂ K) = 1/52

We need to find the probability of the card being spade or ace, i.e.,

P (Spade ‘or’ ace) = P(S ∪ K)

So, by definition of P (A or B) under the axiomatic approach (also called addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (S ∪ K) = P (S) + P (K) – P (S ∩ K)

= ¼ + 1/13 – 1/52

= 17/52 – 1/52

= 16/52

= 4/13

∴ P (S ∪ K) = 4/13

13. The probability that a student will pass the final examination in both English and Hindi is 0.5, and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?

Solution:

Let ‘E’ denotes the event that a student passes in English examination.

And ‘H’ be the event that a student passes in Hindi exam.

It is given that,

P (E) = 0.75

P (passing both) = P (E ∩ H) = 0.5

P (passing neither) = P (E′ ∩ H′) = 0.1

P (H) =?

As, we know that P (A′ ∩ B′) = P (A ∪ B) ′ {using De Morgan’s law}

P (E′ ∩ H′) = P (E ∪ H)′

0.1 = 1 – P (E ∪ H)

P (E ∪ H) = 1 – 0.1

= 0.9

By using the definition of P (A or B) under the axiomatic approach (also called the addition theorem), we know that

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

∴ P (E ∪ H) = P (E) + P (H) – P (E ∩ H)

0.9 = 0.75 + P (H) – 0.5

1.4 – 0.75 = P (H)

∴ P (H) = 0.65

14. One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6.

Solution:

Given: Sample space is the set of the first 100 natural numbers.

n (S) = 100

Let ‘A’ be the event of choosing the number such that it is divisible by 4.

n (A) = [100/4]

= [25]

= 25 {where [.] represents Greatest integer function}

P (A) = n (A) / n (S)

= 25/100

= ¼

Let ‘B’ be the event of choosing the number such that it is divisible by 6.

n (B) = [100/6]

= [16.67]

= 16 {where [.] represents Greatest integer function}

P (B) = n (B) / n (S)

= 16/100

= 4 /25

Now, we need to find the P (such that number chosen is divisible by 4 or 6),

P (A or B) = P (A ∪ B)

By using the definition of P (E or F) under the axiomatic approach (also called the addition theorem), we know that

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

n (A ∩ B) = [100/12]

= [8.33]

= 8

P (A ∩ B) = n (A ∩ B) / n (S)

= 8/100

= 2/25

∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∪ B) = ¼ + 4/25 – 2/25

= ¼ + 2/25

= 33/100

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 33 – Probability

Exercise 33.1 Solutions

Exercise 33.2 Solutions

Exercise 33.3 Solutions

Exercise 33.4 Solutions