NCERT Solution Class 12 Chapter 13 - Probability Exercise 13.4

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.4 – Free PDF Download

Exercise 13.4 of NCERT Solutions for Class 12 Maths Chapter 13 – Probability is based on the following topics:

1. Random Variables and Their Probability Distributions
   1. The probability distribution of a random variable
   2. Mean of a random variable
   3. The variance of a random variable

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.

NCERT Solutions for Class 12 Maths Chapter 13 – Probability Exercise 13.4

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Exercise 13.2 Solutions 18 Questions

Exercise 13.3 Solutions 14 Questions

Exercise 13.5 Solutions 15 Questions

Miscellaneous Exercise On Chapter 13 Solutions 10 Questions

Access Answers to NCERT Class 12 Maths Chapter 13 Exercise 13.4

1. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Solution:

Here we have a table with values for X and P(X).

As we know, the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of the given table = 0.4 + 0.4 + 0.2

= 1

Hence, the given table shows the probability distributions of a random variable.

Solution:

Here we have a table with values for X and P(X).

As we see from the table that P(X) = -0.1 for X = 3.

It is known that the probability of any observation must always be positive and that it can’t be negative.

Hence, the given table does not show the probability distributions of a random variable.

Solution:

Here we have a table with values for X and P(X).

As we know, the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of the given table = 0.6 + 0.1 + 0.2

= 0.9 ≠ 1

Hence, the given table does not show the probability distributions of a random variable.

Solution:

Here we have a table with values for X and P(X).

As we know, the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table = 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05 ≠ 1

Hence, the given table does not show the probability distributions of a random variable.

2. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Solution:

Given urn contains 5 red and 2 black balls.

Let R represent the red balls and B represent the black balls.

Two balls are drawn randomly.

Hence, the sample space of the experiment is S = {BB, BR, RB, RR}

X represents the number of black balls.

⇒ X (BB) = 2

X (BR) = 1

X (RB) = 1

X (RR) = 0

Therefore, X is a function on sample space whose range is {0, 1, 2}.

Thus, X is a random variable which can take the values 0, 1 or 2.

3. Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Solution:

Given a coin is tossed 6 times.

X represents the difference between the number of heads and the number of tails.

⇒ X (6H, 0T) = |6-0| = 6

X (5H, 1T) = |5-1| = 4

X (4H, 2T) = |4-2| = 2

X (3H, 3T) = |3-3| = 0

X (2H, 4T) = |2-4| = 2

X (1H, 5T) = |1-5| = 4

X (0H, 6T) = |0-6| = 6

Therefore, X is a function on sample space whose range is {0, 2, 4, 6}.

Thus, X is a random variable which can take the values 0, 2, 4 or 6.

4. Find the probability distribution of
(i) number of heads in two tosses of a coin.

Solution:

Given a coin is tossed twice.

Hence, the sample space of the experiment is S = {HH, HT, TH, TT}

X represents the number of heads.

⇒ X (HH) = 2

X (HT) = 1

X (TH) = 1

X (TT) = 0

Therefore, X is a function on sample space whose range is {0, 1, 2}.

Thus, X is a random variable which can take the values 0, 1 or 2.

As we know,

P (HH) = P (HT) = P (TH) = P (TT) = 1/4

P (X = 0) = P (TT) = 1/4

P (X = 1) = P (HT) + P (TH) = 1/4 + 1/4 = 1/2

P (X = 2) = P (HH) = 1/4

Hence, the required probability distribution is,

(ii) Number of tails in the simultaneous tosses of three coins.

Solution:

Given three coins are tossed simultaneously. Hence, the sample space of the experiment is S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

X represents the number of tails.

As we see, X is a function on sample space whose range is {0, 1, 2, 3}.

Thus, X is a random variable which can take the values 0, 1, 2 or 3.

P (X = 0) = P (HHH) = 1/8

P (X = 1) = P (HHT) + P (HTH) + P (THH) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 2) = P (HTT) + P (THT) + P (TTH) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 3) = P (TTT) = 1/8

Hence, the required probability distribution is,

(iii) Number of heads in four tosses of a coin.

Solution:

Given four tosses of a coin.

Hence, the sample space of the experiment is

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

X represents the number of heads.

As we see, X is a function on sample space whose range is {0, 1, 2, 3, 4}.

Thus, X is a random variable which can take the values 0, 1, 2, 3 or 4.

P (X = 0) = P (TTTT) = 1/16

P (X = 1) = P (HTTT) + P (TTTH) + P (THTT) + P (TTHT) = 1/16 + 1/16 + 1/16 + 1/16 = ¼

P(X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (THTH) + P (HTHT) + P(HTTH)= 1 /16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8

P(X = 3) = P (THHH) + P (HHHT) + P (HTHH) + P (HHTH) = 1/16 + 1/16 + 1/16 + 1/16 = ¼

P(X = 4) = P (HHHH) = 1/16

Hence, the required probability distribution is,

5. Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

Solution:

Given a die is tossed two times.

When a die is tossed two times, then the number of observations will be (6 × 6) = 36.

Now, let X be a random variable which represents the success.

(i) Here, success is given as a number greater than 4.

Now

P (X = 0) = P (number ≤ 4 in both tosses) = 4/6 × 4/6 = 4/9

P (X = 1) = P (number ≤ 4 in first toss and number ≥ 4 in second case) + P (number ≥ 4 in first toss and number ≤ 4 in second case) is

= (4/6 × 2/6) + (2/6 × 4/6) = 4/9

P (X = 2) = P (number ≥ 4 in both tosses) = 2/6 × 2/6 = 1/9

Hence, the required probability distribution is,

(ii) Here, success is given as six appears on at least one die.

Now P (X = 0) = P (six does not appear on any die) = 5/6 × 5/6 = 25/36

P (X = 1) = P (six appears on at least one of the dies) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18

Hence, the required probability distribution is,

6. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Solution:

Given a lot of 30 bulbs which include 6 defectives.

Then, number of non-defective bulbs = 30 – 6 = 24

4 bulbs are drawn at random with replacement.

Let X denote the number of defective bulbs from the selected bulbs.

Clearly, X can take the value of 0, 1, 2, 3 or 4.

P (X = 0) = P (4 are non-defective and 0 defective)

P (X = 1) = P (3 are non-defective and 1 defective)

P (X = 2) = P (2 are non-defective and 2 defective)

P (X = 3) = P (1 is non-defective and 3 defective)

P (X = 4) = P (0 is non-defective and 4 defective)

Hence, the required probability distribution is,

7. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution:

Given, a head is 3 times as likely to occur as a tail.

Now, let the probability of getting a tail in the biased coin be x.

⇒ P (T) = x

And P (H) = 3x

For a biased coin, P (T) + P (H) = 1

⇒ x + 3x = 1

⇒ 4x = 1

⇒ x = 1/4

Hence, P (T) = 1/4 and P (H) = 3/4

As the coin is tossed twice, the sample space is {HH, HT, TH, TT}

Let X be a random variable representing the number of tails.

Clearly, X can take the value of 0, 1 or 2.

P(X = 0) = P (no tail) = P (H) × P (H) = ¾ × ¾ = 9/16

P(X = 1) = P (one tail) = P (HT) × P (TH) = ¾. ¼ × ¼. ¾ = 3/8

P(X = 2) = P (two tail) = P (T) × P (T) = ¼ × ¼ = 1/16

Hence, the required probability distribution is,

8. A random variable X has the following probability distribution:

Determine

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

Solution:

Given a random variable X with its probability distribution.

(i) As we know, the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence, the sum of probabilities of the given table:

⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7K² + k = 1

⇒ 10K² + 9k = 1

⇒ 10K² + 9k – 1 = 0

⇒ (10K-1) (k + 1) = 0

k = -1, 1/10

It is known that the probability of any observation must always be positive and that it can’t be negative.

So k = 1/10

(ii) Now we have to find P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k

P (X < 3) = 3 × 1/10 = 3/10

(iii) Now we have to find P(X > 6)

P(X > 6) = P(X = 7)

= 7K² + k

= 7 × (1/10)² + 1/10

= 7/100 + 1/10

P (X > 6) = 17/100

(iv) Consider P (0 < X < 3)

P (0 < X < 3) = P(X = 1) + P(X = 2)

= k + 2k

= 3k

P (0 < X < 3) = 3 × 1/10 = 3/10

9. The random variable X has a probability distribution P(X) of the following form, where k is some number:

(a) Determine the value of k.
(b) Find P (X < 2), P (X ≤ 2), P(X ≥ 2).

Solution:

Given: A random variable X with its probability distribution.

(a) As we know, the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of the given table:

⇒ k + 2k + 3k + 0 = 1

⇒ 6k = 1

k = 1/6

(b) Now we have to find P(X < 2)

P (X < 2) = P (X = 0) + P (X = 1)

= k + 2k

= 3k

P (X < 2) = 3 × 1/6 = ½

Consider P (X ≤ 2)

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

= k + 2k + 3k

= 6k

P (X ≤ 2) = 6 × 1/6 = 1

Now we have to find P(X ≥ 2)

P(X ≥ 2) = P(X = 2) + P(X > 2)

= 3k + 0

= 3k

P (X ≥ 2) = 3 × 1/6 = ½

10. Find the mean number of heads in three tosses of a fair coin.

Solution:

Given a coin is tossed three times.

Three coins are tossed simultaneously. Hence, the sample space of the experiment is S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

X represents the number of heads.

As we see, X is a function on sample space whose range is {0, 1, 2, 3}.

Thus, X is a random variable which can take the values 0, 1, 2 or 3.

P (X = 0) = P (TTT) = 1/8

P (X = 1) = P (TTH) + P (THT) + P (HTT) = 1/8 +1/8 + 1/8 = 3/8

P (X = 2) = P (THH) + P (HTH) + P (HHT) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 3) = P (HHH) = 1/8

Hence, the required probability distribution is,

Therefore mean μ is:

11. Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Solution:

Given a die is thrown two times.

When a die is tossed two times, then the number of observations will be (6 × 6) = 36.

Now, let X be a random variable which represents the success and is given as six appears on at least one die.

Now

P (X = 0) = P (six does not appear on any die) = 5/6 × 5/6 = 25/36

P (X = 1) = P (six appears on least one of the dies) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18

P (X = 2) = P (six does appear on both dies) = 1/6 × 1/6 = 1/36

Hence, the required probability distribution is,

Therefore Expectation of X E (X):

12. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

Solution:

Given the first six positive integers.

Two numbers can be selected at random (without replacement) from the first six positive integers in 6 × 5 = 30 ways.

X denote the larger of the two numbers obtained. Hence, X can take any value of 2, 3, 4, 5 or 6.

For X = 2, the possible observations are (1, 2) and (2, 1)

P (X) = 2/30 = 1/15

For X = 3, the possible observations are (1, 3), (3, 1), (2, 3) and (3, 2).

P (X) = 4/30 = 2/15

For X = 4, the possible observations are (1, 4), (4, 1), (2,4), (4,2), (3,4) and (4,3).

P (X) = 6/30 = 1/5

For X = 5, the possible observations are (1, 5), (5, 1), (2,5), (5,2), (3,5), (5,3), (5, 4) and (4,5).

P (X) = 8/30 = 4/15

For X = 6, the possible observations are (1, 6), (6, 1), (2,6), (6,2), (3,6), (6,3), (6, 4), (4,6), (5,6) and (6,5).

P (X) = 10/30 = 1/3

Hence, the required probability distribution is,

Therefore E(X) is:

13. Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Solution:

Given two fair dice are rolled

When two fair dice are rolled, the number of observations will be 6 × 6 = 36.

X denote the sum of the numbers obtained when two fair dice are rolled. Hence, X can take any value of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.

For X = 2, the possible observations are (1, 1).

P (X) = 1/36

For X = 3, the possible observations are (1, 2) and (2, 1)

P (X) = 2/36 = 1/18

For X = 4, the possible observations are (1, 3), (2, 2) and (3, 1).

P (X) = 3/36 = 1/12

For X = 5, the possible observations are (1, 4), (4, 1), (2, 3) and (3, 2)

P (X) = 4/39 = 1/9

For X = 6, the possible observations are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3).

P (X) = 5/36

For X = 7, the possible observations are (1, 6), (6, 1), (2,5), (5,2),(3,4) and (4,3).

P (X) = 6/36 = 1/6

For X = 8, the possible observations are (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4).

P (X) = 5/36

For X = 9, the possible observations are (5, 4), (4, 5), (3, 6) and (6, 3)

P (X) = 4/36 = 1/9

For X = 10, the possible observations are (5, 5), (4, 6) and (6, 4).

P (X) = 3/36 = 1/12

For X = 11, the possible observations are (6, 5) and (5, 6)

P (X) = 2/36 = 1/18

For X = 12, the possible observations are (6, 6).

P (X) = 1/36

Hence, the required probability distribution is,

Therefore E(X) is:

E (X) = 7

= √5.833

= 2.415

14. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Solution:

Given a class of 15 students with their ages.

From the given data, we can draw a table

P(X = 14) = 2/15

P(X = 15) = 1/15

P(X = 16) = 2/15

P(X = 17) = 3/15

P(X = 18) = 1/15

P(X = 19) = 2/15

P(X = 20) = 3/15

P(X = 21) = 1/15

Hence, the required probability distribution is,

Therefore E(X) is:

15. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X).

Solution:

Given: X = 0 if members oppose, and X = 1 if members are in favour.

P(X = 0) = 30% = 30/100 = 0.3

P(X = 1) = 70% = 70/100 = 0.7

Hence, the required probability distribution is,

Therefore E(X) is:

= 0 × 0.3 + 1 × 0.7

⇒ E(X) = 0.7

And E(X²) is:

= (0)² × 0.3 + (1)² × 0.7

⇒ E(X²) = 0.7

Then Variance, Var(X) = E(X²) – (E(X))²

= 0.7 – (0.7)²

= 0.7 – 0.49 = 0.21

16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
A. 1
B. 2
C. 5
D. 8/3

Solution:

B. 2

Explanation:

Given a die having 1 written on three faces, 2 on two faces and 5 on one face.

Let X be the random variable representing a number on the given die.

Then X can take any value of 1, 2 or 5.

The total numbers is six.

Now

P(X = 1) = 3/6 = ½

P(X = 2) = 1/3

P(X = 5) = 1/6

Hence, the required probability distribution is,

Therefore, Expectation of X E(X):

17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
A. 37/221
B. 5/13
C. 1/13
D. 2/13

Solution:

D. 2/13

Explanation:

Given a deck of cards.

X be the number of aces obtained.

Hence, X can take value of 0, 1 or 2.

As we know, in a deck of 52 cards, 4 cards are aces. Therefore 48 cards are non- ace cards.

= 6/1326

Hence, the required probability distribution is,

Therefore Expectation of X E(X):

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