NCERT Solutions for Maths Class 12 Exercise 13.4

NCERT Solutions for Maths Class 12 Exercise 13.4

NCERT Solutions for Maths Class 12 Exercise 13.4

Hello Students. Welcome to maths-formula.com. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 13.4.

You can download the PDF of NCERT Books Maths Chapter 10 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 13.4.

Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

NCERT Solutions for Maths Class 12 Exercise 13.4 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from NCERT Syllabus for Mathematics Class 12.

NCERT Solutions for Maths Class 12 Exercise 13.4 are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

If you want to recall All Maths Formulas for Class 12, you can find it by clicking this link.

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NCERT Solutions for Maths Class 12 Exercise 13.1

NCERT Solutions for Maths Class 12 Exercise 13.2

NCERT Solutions for Maths Class 12 Exercise 13.3

NCERT Solutions for Maths Class 12 Exercise 13.5

NCERT Solutions for Maths Class 12 Exercise 13.4

Maths Class 12 Ex 13.4 Question 1.

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

Solution:

(i) P(0) + P(1) + P(2) = 0.4 + 0.4 + 0.2 = 1

It is a probability distribution.

(ii) P(3) = –0.1 which is not possible.
Hence, it is not a probability distribution.

(iii) P(–1) + P(0) + P(1) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, it is not a probability distribution.

(iv) P(3) + P(2) + P(1) + P(0) + P(–1) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 ≠ 1
Hence, it is not a probability distribution.

Maths Class 12 Ex 13.4 Question 2.

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

Solution:

These two balls may be selected as RR, RB, BR, BB, where R represents red and B represents black ball.

Variable X has the values 0, 1, 2, i.e., there may be no black balls, may be one black ball, or both the balls are black.

Since, X is a number whose values are defined on the outcomes of a random experiment, therefore, X is a random variable.

Maths Class 12 Ex 13.4 Question 3.

Let X represents 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:

A coin is tossed six times.

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

Maths Class 12 Ex 13.4 Question 4.

Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.

Solution:

Maths Class 12 Ex 13.4 Question 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:

Maths Class 12 Ex 13.4 Question 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:

There are 30 bulbs which include 6 defective bulbs.
Probability of getting a defective bulb = 6/30 = 1/5
Probability of getting a good bulb = 1 – 1/5 = 4/5
Let X denotes variable of defective bulbs in a sample of 4 bulbs.

Maths Class 12 Ex 13.4 Question 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:

Let p represents the appearance of tail and q represents the appearance of head.
Now, q = 3p as p + q = 1, we have p + 3p = 1

Maths Class 12 Ex 13.4 Question 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:

(i) Sum of probabilities = 1

Maths Class 12 Ex 13.4 Question 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:

(a) Sum of probabilities = 1
k + 2k + 3k + 0 = 1 or 6k = 1, k = 1/6
The probability distribution is as given below:

Maths Class 12 Ex 13.4 Question 10.

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

Solution:

S = {H, T}, n(S) = 2
Let A denotes the appearance of head on a toss, A = {H}

Maths Class 12 Ex 13.4 Question 11.

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

Solution:

Two dice thrown simultaneously is the same as the die thrown 2 times.
Let S = {1, 2, 3, 4, 5, 6}, n(S) = 6
Let A denotes the number 6.

Maths Class 12 Ex 13.4 Question 12.

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

Solution:

There are six numbers 1, 2, 3, 4, 5, 6. One of them is selected in 6 ways.

When one of the numbers has been selected, 5 numbers are left, one number out of 5 may be select in 5 ways.

Number of ways of selecting two numbers without replacement out of 6 positive integers

= 6 × 5 = 30.

Maths Class 12 Ex 13.4 Question 13.

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

Solution:

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

(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2),

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

n(S) = 36
Let A denotes the sum of the numbers = 2
B denotes the sum of the numbers = 3
C denotes the sum of the numbers = 4
D denotes the sum of the numbers = 5
E denotes the sum of the numbers = 6
F denotes the sum of the numbers = 7
G denotes the sum of the numbers = 8
H denotes the sum of the numbers = 9
I denotes the sum of the numbers = 10
J denotes the sum of the numbers = 11
K denotes the sum of the numbers = 12

Maths Class 12 Ex 13.4 Question 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:

There are 15 students in a class. Each has the same chance of being chosen.
The probability of each student to be selected = 1/15

Maths Class 12 Ex 13.4 Question 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:

Here the variable values are 1 and 0 and the probability of occurrence is 70% = 0.7 and 30% = 0.3
Probability distribution is

Choose the correct answer in each of the following:

Maths Class 12 Ex 13.4 Question 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:

Therefore, mean is 2.
Hence, the correct answer is option (B).

Maths Class 12 Ex 13.4 Question 17.

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

Solution:

n(S) = 52, n(A) = 4

Now, E(X) = 2/13
Hence, the correct answer is option (D).