Ncert Solutions For Class 12 Maths Ex 13.4

Ncert Solutions For Class 12 Maths Chapter 13 Ex 13.4

Q-1: Check whether the following are the probability distributions of a random variable or not. Also, justify your answer.

(i)

X 0 1 2
P(x) 0.3 0.5 0.2

(ii)

X 0 1 2 3 4
P(X) 0.2 0.6 0.3 -0.2 0.3

(iii)

Y 0 1 2
P(Y) 0.7 0.2 0.3

(iv)

Z 3 2 1 0 -1
P(Z) 0.4 0.3 0.5 0.2 0.06

Solution:

We know that the sum of all of the probabilities in the probability distribution should be one.

(i)

X 0 1 2
P(x) 0.3 0.5 0.2

 

Sum of the probabilities = 0.3 + 0.5 + 0.2 = 1.0

Hence, this given table is a probability distribution of the random variables.

 

(ii)

X 0 1 2 3 4
P(X) 0.2 0.6 0.3 -0.2 0.3

 

Here, from the table we can see that for X = 3, P(X) = -0.2

We know that, the probability for any of the observation cannot be negative.

Hence, the table given is not a probability distribution for the random variables.

(iii)

Y 0 1 2
P(Y) 0.7 0.2 0.3

 

Sum of all the probabilities given in the Q. = 0.7 + 0.2 + 0.3 = 1.2 ≠ 1

Hence, the table given is not a probability distribution for the random variables.

 

(iv)

Z 3 2 1 0 -1
P(Z) 0.4 0.3 0.5 0.2 0.06

 

Here, sum of all of the probabilities given in the Q. = 0.4 + 0.3 + 0.5 + 0.2 + 0.06 = 1.46 1

Hence, the table given is not a probability distribution for the random variables.

 

Q.2: There are 5 red balls and 2 black balls. From that, two balls are drawn at random. Let X represents the number of the black balls. Find the values for X. Check whether X is a random variable or not?

Solution:

Let, the two balls drawn randomly are represented as RR, RB, BR, and BB, where R is for a black ball and B is for a red ball.

According to Q,

X represents the number of the black balls.

Thus,

X (RR) = 0

X (RB) = 1

X (BR) = 1

X (BB) = 2

Hence, the possible values of X are 0, 1 and 2.

Therefore, C is a random variable.

 

Q.3: Let us consider X which represents the difference between the total number of tails and the number of heads which can be obtained when a single coin is tossed for 5 times. Find the possible values for X.

Solution:

As per the data given in the Q,

A coin will be tossed for 5 times and every time the result is observed. Also, X represents the difference between the total number of heads and tails observed after each toss.

Thus,

X (5H, 0T) = | 5 – 0 | = 5

X (4H, 1T) = | 4 – 1 | = 3

X (3H, 2 T) = | 3 – 2 | = 1

X(2H, 3T) = | 2 – 3 | = 1

X(1H, 4T) = | 1 – 4 | = 3

X(0H, 5T) = | 0 – 5 | = 5

Hence, the possible values for X are 5, 3 and 1.

 

Q.4: What will be the probability distribution for?

(i) Number of tails after tossing a coin for twice.

(ii) Number of heads after simultaneously tossing a coin for three times.

(iii) Number of tails after tossing a coin for four times.

Solution:

(i) Once a coin is tossed for two times, the chances are:

{TT, TH, HT, HH}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

X(TT) = 2, X(TH) = 1, X(HT) = 1 and X(HH) = 0.

Hence, X can take the values of 0, 1 and 2.

We know that,

P(TT) = P(TH) = P(HT) = P(HH) = \(\frac{ 1 }{ 4 }\)

P(X = 0) = P(HH) = \(\frac{ 1 }{ 4 }\)

P(X = 1) = P (HT) + P(TH) = \(\frac{ 1 }{ 4 } + \frac{ 1 }{ 4 }\) = \(\frac{ 1 + 1 }{ 4 } = \frac{ 2 }{ 4 } = \frac{ 1 }{ 2 }\)

P(X = 2) = P(TT) = \(\frac{ 1 }{ 4 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2
P(X) \(\frac{ 1 }{ 4 }\) \(\frac{ 1 }{ 2 }\) \(\frac{ 1 }{ 4 }\)

(ii) Once a coin is tossed for three times, the chances are:

{TTT, TTH, THH, HHH, HHT, HTT, HTH, THT}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

X(HHH) = 3, X(HTH) = 2, X(HHT) = 2, X(THH) = 2, X(TTH) = 1, X(HTT) = 1, X(THT) = 1 and X(TTT) = 0.

Hence, X can take the values of 0, 1, 2 and 3.

Now,

P(X = 0) = P(TTT) = \(\frac{ 1 }{ 8 }\)

P(X = 1) = P (TTH) + P(HTT) + P(THT) = \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\) = \(\frac{ 1 + 1 + 1 }{ 8 } = \frac{ 3 }{ 8 } \)

P(X = 2) = P (HTH) + P(THH) + P(HHT) = \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\) = \(\frac{ 1 + 1 + 1 }{ 8 } = \frac{ 3 }{ 8 } \)

P(X = 3) = P(HHH) = \(\frac{ 1 }{ 8 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2 3
P(X) \(\frac{ 1 }{ 8 }\) \(\frac{ 3 }{ 8 }\) \(\frac{ 3 }{ 8 }\) \(\frac{ 1 }{ 8 }\)

 

(iii) Once a coin is tossed for four times, the chances are:

{TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHH, HTHT, HHTT, HHTH, HHHT, HHHH}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

X(TTTT) = 4, X(TTTH) = 3, X(TTHT) = 3, X(THTT) = 3, X(HTTT) = 3, X(TTHH) = 2, X(THTH) = 2, X(THHT) = 2, X(HTTH) = 2, X(HTHT) = 2, X(HHTT) = 2, X(THHH) = 1, X(HTHH) = 1, X(HHTH) = 1, X(HHHT) = 1 and X(HHHH) = 0.

Hence, X can take the values of 0, 1, 2, 3 and 4.

Now,

P(X = 0) = P(HHHH) = \(\frac{ 1 }{ 16 }\)

P(X = 1) = P (THHH) + P(HTHH) + P(HHTH) + P(HHHT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 }{ 16 } = \frac{ 4 }{ 16 } = \frac{ 1 }{ 4 } \)

P(X = 2) = P (TTHH) + P(THTH) + P(THHT) + P(HTTH) + P(HTHT) + P(HHTT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 + 1 + 1 }{ 16 } = \frac{ 6 }{ 16 } = \frac{ 3 }{ 8 } \)

P(X = 3) = P (TTTH) + P(TTHT) + P(THTT) + P(HTTT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 }{ 16 } = \frac{ 4 }{ 16 } = \frac{ 1 }{ 4 } \)

P(X = 0) = P(TTTT ) = \(\frac{ 1 }{ 16 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2 3 4
P(X) \(\frac{ 1 }{ 16 }\) \(\frac{ 1 }{ 4 }\) \(\frac{ 3 }{ 8 }\) \(\frac{ 1 }{ 4 }\) \(\frac{ 1 }{ 16 }\)

 

Q.5: What will be the probability distribution of the numbers for the two success tosses of a die, where a success will be defined as?

(i) Obtained number which is greater than 3.

(ii) On the die, six appears for at least once.

Solution:

When we will toss a die for two times, we will obtain at least (6 × 6) = 36 number of the observations.

Let us consider X which is a random variable, which represents total number of successes.

(i) In this case, numbers which is greater than 3 is referred to as success.

P(X = 0) = P (Number which are less than or equal to 3 after having both the tosses) = \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 } = \frac{ 1 }{ 4 }\)

P(X = 1) = P(Numbers which are less than or equal to 3 in first toss and greater than 3 in the second toss) + P(Numbers which are greater than 3 in first toss and less than or equal to 3 in the second toss)

= \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 } + \frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 }\)

= \( \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } + \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \( \frac{ 1 }{ 4 } + \frac{ 1 }{ 4 }\)

= \(\frac{ 1 }{ 2 }\)

P(X = 2) = P(Numbers which are greater than 3 after both toss) = \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 }\)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 4 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2
P(X) \(\frac{ 1 }{ 4 }\) \(\frac{ 1 }{ 2 }\) \(\frac{ 1 }{ 4 }\)

 

(ii) In this case, success will be when six appeared at least for once after tossing the dice simultaneously.

P(Y = 0) = P(six doesn’t appeared on either of the dice) = \(\frac{ 5 }{ 6 } \times \frac{ 5 }{ 6 } = \frac{ 25 }{ 36 } \)

P(Y = 1) = P(six appeared at least on one of the dice) = \(\frac{ 1 }{ 6 } \times \frac{ 5 }{ 6 } + \frac{ 5 }{ 6 } \times \frac{ 1 }{ 6 } = \frac{ 5 }{ 36 } + \frac{ 5 }{ 36 } = \frac{ 10 }{ 36 } = \frac{ 5 }{ 18 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 0 1
P(Y) \(\frac{ 25 }{ 36 }\) \(\frac{ 5 }{ 18 }\)

 

 

Q.6: Consider a lot of 34 bulbs among which 10 bulbs are defective. From that lot, a sample of 3 bulbs is drawn at random with having replacement. What is the probability distribution for the number of the defective bulbs?

Solution:

As per the data given in the Q., we have

A lot of 34 bulbs among which 10 bulbs are defective.

So, the number of non- defective bulbs = 34 – 10 = 24 bulbs

Now,

A sample of 3 bulbs is drawn at random from the lot with having replacement.

Let us consider X be a random variable which denotes the total number of defective bulbs among the sample drawn.

P(X = 0) = P (3 non- defective and 0 defective) = 3C0. \(\frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 0! 3! } \times \frac{ 27 }{ 64 }\) = \(1 \times \frac{ 27 }{ 64 } = \frac{ 27 }{ 64 }\)

P(X = 1) = P (2 non- defective and 1 defective) = 3C1. \(\frac{ 1 }{ 4 } \times \frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 1! 2! } \times \frac{ 9 }{ 64 }\) = \(\frac{ 3 \times 2! }{ 2! } \times \frac{ 9 }{ 64 } = 3 \times \frac{ 9 } { 64 } = \frac{ 27 }{ 64 }\)

P(X = 2) = P (1 non- defective and 2 defective) = 3C2. \(\frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 2! 1! } \times \frac{ 3 }{ 64 }\) = \(\frac{ 3 \times 2! }{ 2! } \times \frac{ 3 }{ 64 } = 3 \times \frac{ 3 } { 64 } = \frac{ 9 }{ 64 }\)

P(X = 3) = P (0 non- defective and 3 defective) = 3C3. \(\frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 }\) = \(\frac{ 3! }{ 3! 0! } \times \frac{ 1 }{ 64 }\) = \(1 \times \frac{ 1 }{ 64 } = \frac{ 1 }{ 64 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2 3
P(X) \(\frac{ 27 }{ 64 }\) \(\frac{ 27 }{ 64 }\) \(\frac{ 9 }{ 64 }\) \(\frac{ 1 }{ 64 }\)

 

 

Q.7: Consider a situation where a coin will be biased in such a manner that the tail will occur likely 4 times the head. What will be the probability distribution of the number of heads, by considering the situation that the coin is tossed for two times?

Solution:

Let us consider that the probability of getting the head in the biased coin is x.

Then,

P(H) = x

⟹ P(T) = 4x

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

⟹ x + 4x = 1

⟹ 5x = 1

⟹ x = \(\frac{ 1 }{ 5 }\)

Hence, P(H) = \(\frac{ 1 }{ 5 }\) and P(T) = \(\frac{ 4 }{ 5 }\)

When the coin will be tossed for two times, then the sample space will be { TT, TH, HT, HH }

Let, X be the random variable which represents the number of heads.

Then,

P(X = 0) = P(no heads) = P(T) × P(T) = \(\frac{ 4 }{ 5 } \times \frac{ 4 }{ 5 } = \frac{ 16 }{ 25 }\)

P(X = 1) = P(one heads) = P(HT) + P(TH) = \(\frac{ 1 }{ 5 } \times \frac{ 4 }{ 5 } + \frac{ 4 }{ 5 } \times \frac{ 1 }{ 5 } = \frac{ 4 }{ 25 } + \frac{ 4 }{ 25 } = \frac{ 8 }{ 25 }\)

P(X = 2) = P(two heads) = P(H) × P(H) = \(\frac{ 1 }{ 5 } \times \frac{ 1 }{ 5 } = \frac{ 1 }{ 25 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2
P(X) \(\frac{ 16}{ 25 }\) \(\frac{ 8 }{ 25 }\) \(\frac{ 1 }{ 25 }\)

Q.8: The following probability distribution is for the random variable X.

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k

Find:

(i) The value of k

(ii) P(X < 4)

(iii) P(X > 5)

(iv) P (0 < X < 2)

Solution:

We know that, the sum of all of the probabilities of the probability distribution of a random variable is 1.

Thus,

k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1

⟹ 10k2 + 9k = 1

⟹ 10k2 + 9k – 1 = 0

⟹ 10k2 + 10k – k – 1 = 0

⟹ 10k(k + 1) – 1(k + 1) = 0

⟹ (k + 1)(10k – 1) = 0

⟹ k = -1, \(\frac{ 1 }{ 10 }\)

Now,

k = -1 is not possible because the probability for any of the event cannot ever be negative.

So, k = \(\frac{ 1 }{ 10 }\)

 

(ii) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= o + k + 2k + 2k = 5k

= \(5 \times \frac{ 1 }{ 10 }\) = \(\frac{ 1 }{ 2 }\)

 

(iii) P(X > 5) = P(X = 6) + P(X = 7)

= 2k2 + 7k2 + k

= 9k2 + k

= k (9k + 1)

= \(\frac{ 1 }{ 10 }\)(\( 9 \times \frac{ 1 }{ 10 }\) + 1)

= \(\frac{ 1 }{ 10 }\)(\(\frac{ 9 }{ 10 } + 1 \))

= \(\frac{ 1 }{ 10 }\) (\( \frac{ 9 + 10 }{ 10 }\))

= \(\frac{ 1 }{ 10 }\)(\(\frac{ 19 }{ 10 }\))

= \(\frac{ 19 }{ 100 }\)

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

= k = \(\frac{ 1 }{ 10 }\)

 

Q.9: Consider P(Y) be the probability distribution of the random variable Y for the following forms, where some number is.

P(Y) =

2a, if a = 0

3a, if a = 1

4a, if a = 2

0, otherwise

(i) Find the value of a.

(ii) Determine P(Y < 2), P(Y ≥ 2), and P (Y ≤ 2).

Solution:

(i) As we know that the sum for all of the probabilities of the probability distribution of random variables is given by 1.

Thus,

2a + 3a + 4a + 0 = 1

⟹ 9a = 1

⟹ a = \(\frac{ 1 }{ 9 }\)

(ii) P(Y < 2) = P(Y = 0) + P(Y = 1)

= 2a + 3a

= 5a = \( 5 \times \frac{ 1 }{ 9 }\) = \(\frac{ 5 }{ 9 }\)

P(Y ≥ 2) = P(Y = 2) + P(Y > 2)

= 4a + 0

= 4a = \(4 \times \frac{ 1 }{ 9 }\) = \(\frac{ 4 }{ 9 }\)

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

= 4a + 3a + 2a

= 9a = \(9 \times \frac{ 1 }{ 9 }\) = 1

 

 

Q.10: After tossing a fair coin, what will be the mean number for tails?

Solution:

Let us consider X which denotes the chances of success to get tails.

Thus,

For this case, the sample space will be

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

Let, the total number of tails be represented by X.

Then,

X(HHH) = 0, X(HTH) = 1, X(HHT) = 1, X(THH) = 1, X(TTH) = 2, X(HTT) = 2, X(THT) = 2 and X(TTT) = 3.

Here, we can see that X will take values 0, 1, 2 or 3.

Now,

P(X = 0) = P(HHH)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 }\)

P(X = 1) = P(HTH) + P(HHT) + P(THH)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 }\)

P(X = 2) = P(TTH) + P(HTT) + P(THT)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 }\)

P(X = 0) = P(TTT)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) = \(\frac{ 1 }{ 8 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1 2 3
P(X) \(\frac{ 1 }{ 8 }\) \(\frac{ 3 }{ 8 }\) \(\frac{ 3 }{ 8 }\) \(\frac{ 3 }{ 8 }\)

 

Mean of X E(X), µ = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(0 \times \frac{ 1 }{ 8 } + 1 \times \frac{ 3 }{ 8 } + 2 \times \frac{ 3 }{ 8 } + 3 \times \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 } + \frac{ 6 }{ 8 } + \frac{ 3 }{ 8 }\)

= \(\frac{ 12 }{ 8 }\)

= \(\frac{ 3 }{ 2 }\) = 1.5

 

Q.11: Take two dices which are thrown simultaneously. If Y denotes the number of times we will get sixes, then what will be the expectation of Y?

Solution:

As per the data given in the Q., Y represents the total number of sixes which will be obtained after throwing two dice simultaneously.

Thus, Y will take the values of 0, 1 or 2.

Hence,

P(Y = 0) = P (no sixes will be obtained in either of the throw) = \(\frac{ 5 }{ 6 } \times \frac{ 5 }{ 6 } = \frac{ 25 }{ 36 }\)

P(Y = 1) = P (six at the first dice and other number rather than 6 on the second dice) + P(other number rather than six on the first dice and six on the second dice)

= 2 \(\left (\frac{ 5 }{ 6 } + \frac{ 1 }{ 6 } \right)\)

= \( 2 \times \frac{ 5 }{ 36 }\)

= \(\frac{ 5 }{ 18 }\)

P(Y = 2) = P (sixes will be obtained in both of the throw) = \(\frac{ 1 }{ 36 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 0 1 2
P(Y) \(\frac{ 25}{ 36 }\) \(\frac{ 10 }{ 36 }\) \(\frac{ 1 }{ 36 }\)

 

Therefore,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(0 \times \frac{ 25 }{ 36 } + 1 \times \frac{ 10 }{ 36 } + 2 \times \frac{ 1 }{ 36 }\)

= \(0 + \frac{ 5 }{ 18 } + \frac{ 1 }{ 18 }\)

= \(\frac{ 6 }{ 18 }\) = \(\frac{ 1 }{ 3 }\)

 

Q.12: From the starting six positive integers, two numbers will be selected at random, without any replacement. Let, Y denotes the larger number among those two numbers obtained. Find E(Y).

Solution:

As per the Q.’s demand, from the starting six positive integers, two numbers will be selected, without having replacement which will be done in 6 × 5 = 30 ways.

Let us consider Y which represents the two numbers obtained which are larger. Thus,

Y will take values 2, 3, 4, 5 or 6.

For Y = 2, the possible observations will be (1, 2) and (2, 1).

Then,

P(Y = 2) = \(\frac{ 2 }{ 30 } = \frac{ 1 }{ 15 }\)

For Y = 3, the possible observations will be (1, 3), (2, 3), (3, 1) and (3, 2)

Then,

P(Y = 3) = \(\frac{ 4 }{ 30 } = \frac{ 2 }{ 15 }\)

For Y = 4, the possible observations will be (1, 4), (2, 4), (3, 4), (4, 1), (4, 2) and (4, 3)

Then,

P(Y = 4) = \(\frac{ 6 }{ 30 } = \frac{ 1 }{ 5 }\)

For Y = 5, the possible observations will be (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), (5, 1)

Then,

P(Y = 5) = \(\frac{ 8 }{ 30 } = \frac{ 4 }{ 15 }\)

For Y = 6, the possible observations will be (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)

Then,

P(Y = 6) = \(\frac{ 10 }{ 30 } = \frac{ 1 }{ 3 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 2 3 4 5 6
P(X) \(\frac{ 1 }{ 15 }\) \(\frac{ 2 }{ 15 }\) \(\frac{ 1 }{ 5 }\) \(\frac{ 4 }{ 15 }\) \(\frac{ 1 }{ 3 }\)

 

Therefore,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(2 \times \frac{ 1 }{ 15 } + 3 \times \frac{ 2 }{ 15 } + 4 \times \frac{ 1 }{ 5 } + 5 \times \frac{ 4 }{ 15 } + 6 \times \frac{ 1 }{ 3 }\)

= \(\frac{ 2 }{ 15 } + \frac{ 6 }{ 15 } + \frac{ 4 \times 3 }{ 15 } + \frac{ 20 }{ 15 } + \frac{ 6 \times 5 }{ 15 } \)

= \(\frac{ 2 }{ 15 } + \frac{ 6 }{ 15 } + \frac{ 12 }{ 15 } + \frac{ 20 }{ 15 } + \frac{ 30 }{ 15 } \)

= \(\frac{ 70 }{ 15 }\) = \(\frac{ 14 }{ 3 }\)

 

Q.13: Y denotes the sum of all the numbers which are obtained when two fair dice are rolled. What will be the variance and the standard deviation of Y.?

Solution:

Number of observations which are obtained when two dice is rolled is 6 × 6 = 36

P(Y = 2) = P(1, 1) = \(\frac{ 1 }{ 36 }\)

P(Y = 3) = P(1, 2) + P(2, 1) = \(\frac{ 2 }{ 36 }\)

P(Y = 4) = P(1, 3) + P(2, 2) + P(3, 1) = \(\frac{ 3 }{ 36 }\)

P(Y = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(1, 4) = \(\frac{ 4 }{ 36 }\)

P(Y = 6) = P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = \(\frac{ 5 }{ 36 }\)

P(Y = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = \(\frac{ 6 }{ 36 }\)

P(Y = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = \(\frac{ 5 }{ 36 }\)

P(Y = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(3, 6) = \(\frac{ 4 }{ 36 }\)

P(Y = 10) = P(4, 6) + P(5, 5) + P(6, 4) = \(\frac{ 3 }{ 36 }\)

P(Y = 11) = P(5, 6) + P(6, 5) = \(\frac{ 2 }{ 36 }\)

P(Y = 12) = P(6, 6) = \(\frac{ 1 }{ 36 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 2 3 4 5 6 7 8 9 10 11 12
P(Y) \(\frac{ 1 }{ 36 }\) \(\frac{ 2 }{ 36 }\) \(\frac{ 3 }{ 36 }\) \(\frac{ 4 }{ 36 }\) \(\frac{ 5 }{ 36 }\) \(\frac{ 6 }{ 36 }\) \(\frac{ 5 }{ 36 }\) \(\frac{ 4 }{ 36 }\) \(\frac{ 3 }{ 36 }\) \(\frac{ 2 }{ 36 }\) \(\frac{ 1 }{ 36 }\)

 

Therefore,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(2 \times \frac{ 1 }{ 36 } + 3 \times \frac{ 2 }{ 36 } + 4 \times \frac{ 3 }{ 36 } + 5 \times \frac{ 4 }{ 36 } + 6 \times \frac{ 5 }{ 36 } + 7 \times \frac{ 6 }{ 36 } + 8 \times \frac{ 5 }{ 36 } + 9 \times \frac{ 4 }{ 36 } + 10 \times \frac{ 3 }{ 36 } + 11 \times \frac{ 2 }{ 36 } + 12 \times \frac{ 1 }{ 36 } \)

= \(\frac{ 2 }{ 36 } + \frac{ 6 }{ 36 } + \frac{ 12 }{ 36 } + \frac{ 20 }{ 36 } + \frac{ 30 }{ 36 } + \frac{ 42 }{ 36 } + \frac{ 40 }{ 36 } + \frac{ 36 }{ 36 } + \frac{ 30 }{ 36 } + \frac{ 22 }{ 36 } + \frac{ 12 }{ 36 }\)

= \(\frac{ 252 }{ 36 }\) = 7

Expectation of X2 = E(X2) = \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)

= \(4 \times \frac{ 1 }{ 36 } + 9 \times \frac{ 2 }{ 36 } + 16 \times \frac{ 3 }{ 36 } + 25 \times \frac{ 4 }{ 36 } + 36 \times \frac{ 5 }{ 36 } + 49 \times \frac{ 6 }{ 36 } + 64 \times \frac{ 5 }{ 36 } + 81 \times \frac{ 4 }{ 36 } + 100 \times \frac{ 3 }{ 36 } + 121 \times \frac{ 2 }{ 36 } + 144 \times \frac{ 1 }{ 36 } \)

= \(\frac{ 4 }{ 36 } + \frac{ 18 }{ 36 } + \frac{ 48 }{ 36 } + \frac{ 100 }{ 36 } + \frac{ 180 }{ 36 } + \frac{ 294 }{ 36 } + \frac{ 320 }{ 36 } + \frac{ 324 }{ 36 } + \frac{ 300 }{ 36 } + \frac{ 242 }{ 36 } + \frac{ 144 }{ 36 }\)

= \(\frac{ 1974 }{ 36 }\) = 54.8333

Then,

Var(X) = E(X2) – [ E(X)2 ]

= 54.8333 – 49 = 5.8333

Hence,

Standard derivation = \(\sqrt{Var\left (X \right)}\)

= \(\sqrt{ 5.8333 }\) = 2.415

 

Q.14: There is a class having 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student will be selected in such a way that each of them have the same chance of being chosen and Y is recorded as the age of the selected student. Find the probability distribution for the random variable Y. What will be the mean, variance and standard derivation of Y?

Solution:

A class has 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years.

Every student has the same chance of being selected.

Hence, the probability for every student being selected is \(\frac{ 1 }{ 15 }\)

The above information will be compiled in the frequency table which is given below:

Y 14 15 16 17 18 19 20 21
f 2 1 2 3 1 2 3 1

 

P(X = 14) = \(\frac{ 2 }{ 15 }\)

P(X = 15) = \(\frac{ 1 }{ 15 }\)

P(X = 16) = \(\frac{ 2 }{ 15 }\)

P(X = 17) = \(\frac{ 3 }{ 15 }\)

P(X = 18) = \(\frac{ 1 }{ 15 }\)

P(X = 19) = \(\frac{ 2 }{ 15 }\)

P(X = 20) = \(\frac{ 3 }{ 15 }\)

P(X = 21) = \(\frac{ 1 }{ 15 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 14 15 16 17 18 19 20 21
P(Y) \(\frac{ 2 }{ 15 }\) \(\frac{ 1 }{ 15 }\) \(\frac{ 2 }{ 15 }\) \(\frac{ 3 }{ 15 }\) \(\frac{ 1 }{ 15 }\) \(\frac{ 2 }{ 15 }\) \(\frac{ 3 }{ 15 }\) \(\frac{ 1 }{ 15 }\)

 

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(14 \times \frac{ 2 }{ 15 } + 15 \times \frac{ 1 }{ 15 } + 16 \times \frac{ 2 }{ 15 } + 17 \times \frac{ 3 }{ 15 } + 18 \times \frac{ 1 }{ 15 } + 19 \times \frac{ 2 }{ 15 } + 20 \times \frac{ 3 }{ 15 } + 21 \times \frac{ 1 }{ 15 }\)

= \(\\\frac{ 28 }{ 15 } + \frac{ 15 }{ 15 } + \frac{ 32 }{ 15 } + \frac{ 51 }{ 15 } + \frac{ 18 }{ 15 } + \frac{ 38 }{ 15 } + \frac{ 60 }{ 15 } + \frac{ 21 }{ 15 }\)

= \(\frac{ 263 }{ 15 }\) = 17.53

Expectation of X2 = E(X2) = \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)

= \(\left (14 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (15 \right)^{ 2 } \times \frac{ 1 }{ 15 } + \left (16 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (17 \right)^{ 2 } \times \frac{ 3 }{ 15 } + \left (18 \right)^{ 2 } \times \frac{ 1 }{ 15 } + \left (19 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (20 \right)^{ 2 } \times \frac{ 3 }{ 15 } + \left (21 \right)^{ 2 } \times \frac{ 1 }{ 15 }\)

= \(\frac{ 392 }{ 15 } + \frac{ 225 }{ 15 } + \frac{ 512 }{ 15 } + \frac{ 867 }{ 15 } + \frac{ 324 }{ 15 } + \frac{ 722 }{ 15 } + \frac{ 1200 }{ 15 } + \frac{ 441 }{ 15 }\)

= \(\frac{ 4683 }{ 15 }\) = 312.2

Then,

Var(X) = E(X2) – [ E(X)2 ]

= 312.2 – (17.53)2 = 4.78

Hence,

Standard derivation = \(\sqrt{Var\left (X \right)}\)

= \(\sqrt{ 4.78 }\) = 2.186

 

Q.15: 70% members are in the favour and 40% are opposing a certain proposal in a meeting. A member will be randomly selected and we will take Y = 0, if the person opposed, and X = 1 if the person favours. What will be E(Y) and var(Y) in such a situation?

Solution:

As per the data given in the Q., we have

P(Y = 0) = 40% = \(\frac{ 40 }{ 100 }\) = 0.4

P(Y = 1) = (100 – 40)% = 60% = \(\frac{ 60 }{ 100 }\) = 0.6

Hence, the probability distribution which is required as per the Q.’s demand is given below:

X 0 1
P(X) 0.4 0.6

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= 0 × 0.4 + 1 × 0.6 = 0.6

Expectation of X2 = E(X2) = \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)

= 02 × 0.4 + 12 × 0.6 = 0.6

We know that,

Var(X) = E(X2) – [ E(X)2 ]

= 0.6 – (0.6)2 = 0.6 – 0.36 = 0.24

 

Q.16: The mean of the numbers which will be obtained after throwing a die on which 1 is written on two of the faces, 2 on three faces and 5 on one face is

(a) 1

(b) 2

(c) 5

(d) \(\frac{ 13 }{ 6 }\)

Solution:

Let us consider Y which is a random variable representing any number on the die.

Since, it’s a die, then the total number of observations will be six.

Thus, P(Y = 1) = \(\frac{ 2 }{ 6 } = \frac{ 1 }{ 3 }\)

P(Y = 2) = \(\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\)

P(Y = 5) = \(\frac{ 1 }{ 6 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 1 2 5
P(Y) \(\frac{ 1 }{ 3 }\) \(\frac{ 1 }{ 2 }\) \(\frac{ 1 }{ 6 }\)

 

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(1 \times \frac{ 1 }{ 3 } + 2 \times \frac{ 1 }{ 2 } + 5 \times \frac{ 1 }{ 6 }\)

= \(\frac{ 1 }{ 3 } + 1 + \frac{ 5 }{ 6 }\) = \(\frac{ 2 + 6 + 5 }{ 6 }\) = \(\frac{ 13 }{ 6 }\)

Hence, the correct answer is (d).

 

Q.17: Let Y be the number of aces obtained. Let any of the two cards are drawn at random from the deck of the cards. Then, the value of E(X) will be

(a) \(\frac{ 37 }{ 221 }\)

(b) \(\frac{ 5 }{ 13 }\)

(c) \(\frac{ 2 }{ 13 }\)

(d) \(\frac{ 1 }{ 13 }\)

Solution:

Let, Y be the number of aces obtained.

Therefore, Y will take any of the values of 0, 1, or 2

As we know, there are 52 cards in a deck. Among them 4 cards are aces.

Hence, there are 48 non- ace cards.

P(Y = 0) = P(2 non- ace cards and 0 ace card) = \(\frac{ ^{ 48 }C_{ 2 } \times ^{ 4 }C_{ 0 } }{ ^{ 52 }C_{ 2 } } = \frac{ 1128 }{ 1326 }\)

P(Y = 1) = P(1 non- ace cards and 1 ace card) = \(\frac{ ^{ 48 }C_{ 1 } \times ^{ 4 }C_{ 1 } }{ ^{ 52 }C_{ 2 } } = \frac{ 192 }{ 1326 }\)

P(Y = 2) = P(0 non- ace cards and 2 ace card) = \(\frac{ ^{ 48 }C_{ 0 } \times ^{ 4 }C_{ 2 } }{ ^{ 52 }C_{ 2 } } = \frac{ 6 }{ 1326 }\)

Hence, the probability distribution which is required as per the Q.’s demand is given below:

Y 0 1 2
P(Y) \(\frac{ 1128 }{ 1326 }\) \(\frac{ 192 }{ 1326 }\) \(\frac{ 6 }{ 1326 }\)

 

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(0 \times \frac{ 1128 }{ 1326 } + 1 \times \frac{ 192 }{ 1326 } + 2 \times \frac{ 6 }{ 1326 }\)

= \(\frac{ 192 }{ 1326 } + \frac{ 12 }{ 1326 }\) = \(\frac{ 192 + 12 }{ 1326 }\) = \(\frac{ 204 }{ 1326 }\) = \(\frac{ 2 }{ 13 }\)

Hence, the correct answer is (c).