The Exercise 13.4 of NCERT Solutions for Class 12 Maths Chapter 13 â€“ Probability is based on the following topics:

- Random Variables and its Probability Distributions
- Probability distribution of a random variable
- Mean of a random variable
- 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.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 13- Probability Exercise 13.4

### Access other exercises of Class 12 Maths Chapter 13

Exercise 13.1 Solutions 17 Questions

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 of Maths NCERT Class 12 Chapter 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 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 = 0.4 + 0.4 + 0.2

= 1

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

**Solution:**

Here we have 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 probability of any observation must always be positive that it canâ€™t be negative.

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

**Solution:**

Here we have 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 = 0.6 + 0.1 + 0.2

= 0.9 â‰ 1

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

**Solution:**

Here we have 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 â‰ 1

Hence, the given table is not 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 containing 5 red and 2 black balls.

Let R represent red ball and B represent black ball.

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.

â‡’Â 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.

â‡’Â 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.

â‡’Â 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,

X |
0 |
1 |
2 |

P (X) |
1/4 |
1/2 |
1/4 |

**(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)Â = 1/8

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

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

P (X = 3) = P (TTT)Â = 1/8

Hence, the required probability distribution is,

X |
0 |
1 |
2 |
3 |

P (X) |
1/8 |
3/8 |
3/8 |
1/8 |

**(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)Â = 1/16

P (X = 1) = P (HTTT) + P (TTTH) + P (THTT) + P (TTHT)Â = 1/16 + 1/16 + 1/16 + 1/16 = Â¼

P(X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (THTH) + P (HTHT) + P(HTTH)= 1Â /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)Â = 1/16 + 1/16 + 1/16 + 1/16 = Â¼

P(X = 4) = P (HHHH)Â = 1/16

Hence, the required probability distribution is,

X |
0 |
1 |
2 |
3 |
4 |

P (X) |
1/16 |
1/4 |
3/8 |
1/4 |
1/16 |

**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 Ã— 6) = 36.

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

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

Now

P (X = 0) = P (number â‰¤ 4 in both tosses)Â = 4/6 Ã— 4/6 = 4/9Â

P (X = 1) = P (number â‰¤ 4 in first toss and number â‰¥ 4 in second case) + P (number â‰¥ 4 in first toss and number â‰¤ 4 in second case) is

= (4/6 Ã— 2/6) + (2/6 Ã— 4/6) = 4/9

P (X = 2) = P (number â‰¥ 4 in both tosses) = 2/6 Ã— 2/6 = 1/9Â

Hence, the required probability distribution is,

X |
0 |
1 |
2 |

P (X) |
4/9 |
4/9 |
1/9 |

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

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

P (X = 1) = P (six appears at least once of the die) = (1/6 Ã— 5/6) + (5/6 Ã— 1/6) = 10/36 = 5/18

Hence, the required probability distribution is,

X |
0 |
1 |

P (X) |
25/36 |
5/18 |

**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 â€“ 6 = 24

As 4 bulbs are drawn at random with replacement.

Let X denotes 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)Â

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

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

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

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

Hence, the required probability distribution is,

X |
0 |
1 |
2 |
3 |
4 |

P (X) |
256/625 |
256/625 |
96/625 |
16/625 |
1/625 |

**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 head is 3 times as likely to occur as tail.

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

â‡’Â P (T) = x

And P (H) = 3x

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

â‡’Â x + 3x = 1

â‡’Â 4x = 1

â‡’Â x = 1/4

Hence,Â P (T) = 1/4 and P (H) = 3/4

As the coin is tossed twice, so 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) Ã— P (H) = Â¾ Ã— Â¾ = 9/16Â

P(X = 1) = P (one tail) = P (HT) Ã— P (TH) = Â¾. Â¼ Ã— Â¼. Â¾ = 3/8 Â

P(X = 2) = P (two tail) = P (T) Ã— P (T)Â = Â¼ Ã— Â¼ = 1/16

Hence, the required probability distribution is,

X |
0 |
1 |
2 |

P (x) |
9/16 |
3/8 |
1/16 |

**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 given table:

â‡’Â 0 + k + 2k + 2k + 3k + k^{2}Â + 2k^{2}Â + 7K^{2}Â + k = 1

â‡’Â 10K^{2}Â + 9k = 1

â‡’Â 10K^{2}Â + 9k â€“ 1 = 0

â‡’Â (10K-1) (k + 1) = 0

k = -1, 1/10

It is known that probability of any observation must always be positive that it canâ€™t be negative.

So k = 1/10Â

(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 Ã— 1/10 = 3/10

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

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

=Â 7K^{2}Â + k

= 7 Ã— (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 Ã— 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 â‰¤ 2), P(X â‰¥ 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 given table:

â‡’Â k + 2k + 3k + 0 = 1

â‡’Â 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 Ã— 1/6 = Â½

Consider P (X â‰¤ 2)

P (X â‰¤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

= k + 2k + 3k

= 6k

P (X â‰¤ 2) = 6 Ã— 1/6 = 1

Now we have to find P(X â‰¥ 2)

P(X â‰¥ 2) = P(X = 2) + P(X > 2)

=Â 3k + 0

= 3k

P (X â‰¥ 2) = 3 Ã— 1/6 = Â½

**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)Â = 1/8

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

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

P (X = 3) = P (HHH) = 1/8Â

Hence, the required probability distribution is,

X |
0 |
1 |
2 |
3 |

P (X) |
1/8 |
3/8 |
3/8 |
1/8 |

Therefore mean Î¼ 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 Ã— 6) = 36.

Now, let X is 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 of die) = 5/6 Ã— 5/6 = 25/36Â

P (X = 1) = P (six appears at least once of the die) = (1/6 Ã— 5/6) + (5/6 Ã— 1/6) = 10/36 = 5/18 Â

P (X = 2) = P (six does appear on both of die)Â = 1/6 Ã— 1/6 = 1/36

Hence, the required probability distribution is,

X |
0 |
1 |
2 |

P (X) |
25/36 |
5/18 |
1/36 |

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 first six positive integers.

Two numbers can be selected at random (without replacement) from the first six positive integer in 6 Ã— 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,

X |
2 |
3 |
4 |
5 |
6 |

P (X) |
1/15 |
2/15 |
1/5 |
4/15 |
1/3 |

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 then number of observations will be 6 Ã— 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,

X |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |

P (X) |
1/36 |
1/18 |
1/12 |
1/9 |
5/36 |
1/6 |
5/36 |
1/9 |
1/12 |
1/18 |
1/36 |

Therefore E(X) is:

E (X) = 7

**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 the class of 15 students with their ages.

Form the given data we can draw a table

X |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |

f |
2 |
1 |
2 |
3 |
1 |
2 |
3 |
1 |

P(X = 14)Â = 2/15

P(X = 15)Â = 1/15

P(X = 16)Â = 2/15

P(X = 17)Â = 3/15

P(X = 18)Â = 1/15

P(X = 19)Â = 2/15

P(X = 20)Â = 3/15

P(X = 21)Â = 1/15

Hence, the required probability distribution is,

X |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |

P (X) |
2/15 |
1/15 |
2/5 |
3/15 |
1/15 |
2/15 |
3/15 |
1/15 |

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 = 0.3Â

P(X = 1)Â = 70% = 70/100 = 0.7

Hence, the required probability distribution is,

X |
0 |
1 |

P (X) |
0.3 |
0.7 |

Therefore E(X) is:

= 0 Ã— 0.3 + 1 Ã— 0.7

â‡’Â E(X) = 0.7

And E(X^{2}) is:

= (0)^{2}Â Ã— 0.3 + (1)^{2}Â Ã— 0.7

â‡’Â E(X^{2}) = 0.7

Then Variance, Var(X) = E(X^{2}) â€“ (E(X))^{2}

= 0.7 â€“ (0.7)^{2}

= 0.7 â€“ 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 isA. 1B. 2C. 5D. 8/3**

**Solution:**

B. 2

**Explanation:**

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

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

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

The total numbers is six.

Now

P(X = 1) = 3/6 = Â½ Â

P(X = 2)Â = 1/3

P(X = 5)Â = 1/6

Hence, the required probability distribution is,

X |
1 |
2 |
5 |

P (X) |
1/2 |
1/3 |
1/6 |

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) isA.Â 37/221B. 5/13C.Â 1/13D.Â 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,

X |
0 |
1 |
2 |

P (X) |
1128/1326 |
192/1326 |
6/1326 |

Therefore Expectation of X E(X):