Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 13 Probability Ex 13.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 13 Probability Ex 13.4

Question 1.

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

Solution:

(i) Sum of probabilities = 0.4 + 0.4 + 0.2 = 1.

∴ This distribution is a probability deistribution.

(ii) One of the probabilities – 0.1 is negative.

∴ It is not a probability distribution.

(iii) Sum of probabilities is 0.6 + 0.1 + 0.1 = 0.9 ≠ 1.

∴ The given distribution is not a probability distribution.

(iv) Sum of probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.5 = 1.05 > 1

∴ It is not a probability distribution.

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, if X is a random variable?

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 value 0,1,2, i.e., there may be no black balls, may be one black ball, or both the balls are black.

Yes, X is a random variable.

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 the possible values of X?

Solution:

When a coin is tossed 6 times, we obtain the number of heads and tails as follows:

Corresponding values of X are shown here.

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:

(i) When two tosses of a coin are there:

Sample space = {TT, TH, HT, HH}

Zero success ⇒ No head ⇒ Two tails (TT)

∴ P(0) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\).

One success ⇒ 1 head or 1 tail (TH, HT)

P(1) = \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\)

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

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

Two success ⇒ Both heads (HH)

P(2) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\).

∴ Probability Distribution

(ii) When three coins are tossed simultaneously:

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

Zero success = No tail = All head {HHH}

∴ P(0) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\).

1 success = 1 tail and 2 heads

= {THH, HTH, HHT}

∴ P(1) = 3 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\).

1 success = 1 tail and 2 heads

= {THH, HTH, HHT}

2 success = 2 tail and 1 head

= {TTH, THT, HTT}

P(2) = 3 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\).

3 success = All the tails = {TTT}

P(3) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\).

∴ Probability distribution is:

(iii) When a coin is tossed 4 times:

Zero success = No head = All are tails {TTTT}

= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{16}\).

1 success = 1 head 3 tails

= {HTTT, THTT, TTHT, TTTH}

P(1) = ^{4}C_{2} × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{4}{16}\) = \(\frac{1}{4}\).

2 success = 2 heads and 2 tails

= {HHTT, HTHT, HTTH, THHT, THTH, TTHH}

∴ P(2) = ^{4}C_{2} × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{6}{16}\) = \(\frac{3}{8}\).

3 successses = 3 heads and 1 tail

= {HHHT, HHTH, HTHH, THHH}

∴ P(3) = ^{4}C_{3} × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{4}{16}\) = \(\frac{1}{4}\).

4 successes = All the heads = {HHHH}

P(4) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{16}\).

Thus, the probability distribution is:

Question 5.

Find the probability distribution of the number of successses in two tosses of a dice, where a success is defined as

(i) number greater than 4.

(ii) six appeared on at least one dice.

Solution:

(i) When a number on each dice is 1, 2, 3, 4.

Such cases are: {(1, 1), (1, 2), ……………, (2, 1), (2, 2), ……………., (4, 4))} = 16 cases

Total number of cases in sample space

= 6 × 6 = 36.

∴ P(0) = \(\frac{16}{36}\) = \(\frac{4}{9}\).

1 success = 5 or 6 on one dice and 1, 2, 3, 4 on the other dice = 16

∴ P(1) = \(\frac{16}{36}\) = \(\frac{4}{9}\).

Two successes = 5 or 6 on both dice

= (5, 6), (6, 5), (5, 5), (6, 6)

∴ P(2) = \(\frac{4}{36}\) = \(\frac{1}{9}\).

∴ Probability distribution is as follows:

(ii) Zero success = 6 does not appear on any dice.

= 1, 2, 3, 4, 5 appear on both sides

∴ Number of such cases = 25

P(0) = \(\frac{25}{36}\).

1 success ⇒ 6 appears on one dice and 1, 2, 3, 4, 5 appears on both dice

∴ Number of such cases = 25

P(0) = \(\frac{25}{36}\).

1 success ⇒ 6 appears on one dice and 1, 2, 3, 4, 5 appears on the other dice.

Number of cases = 10.

2 successes ⇒ 6 appears on both dice {6, 6}

Number of such cases = 1.

Number of cases, when there is at least one ‘six’ appears

= 10 + 1 = 11.

∴ Probability of getting at least 1 six = \(\frac{11}{36}\).

∴ Probability distribution is:

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 number of defective bulbs.

Solution:

There are 30 bulbs, which include 6 defective bulbs.

Probability of getting a defective bulb = \(\frac{6}{30}\) = \(\frac{1}{5}\).

Probability of getting a good bulb = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\).

Let X denotes variable of defective bulbs in a sample of 4 bulbs.

∴ Probability distribution of defective bulbs is:

Question 7.

A com 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 tails.

Solution:

When the coin is tossed, the head occurs 3 times the tail occurs.

Let the tail occurs x times.

∴ Head occurs 3x times.

So, total number of cases = x + 3x = 4x.

∴ Probability of getting a head = \(\frac{3x}{4x}\) = \(\frac{3}{4}\).

∴ Probability of getting a tail = \(\frac{x}{4x}\) = \(\frac{1}{4}\).

When no tails occurs, the probability of getting {HH}

= \(\frac{3}{4}\) × \(\frac{1}{4}\) + \(\frac{1}{4}\) × \(\frac{3}{4}\) = \(\frac{3}{16}\) × 2 = \(\frac{3}{8}\).

The probability of getting both tails

P(2) = \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\).

∴ Probability distribution of occuring tails:

Question 8.

A random variable X has the following probability distribution:

Determine:

(i) k’

(ii) P(X< 3) (iii) P(X > 0)

(iv) P(0 < X < 3)

Solution:

(i) Sum of probabilities = 1

i.e; o + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2} + k = 1

or 10k^{2} + 9k = 1 or 10k^{2} + 9k – 1 = 0

⇒ (k + 1)(10k – 1) = 0

∴ k = – 1 or k = \(\frac{1}{10}\)

k ≠ – 1 ∴ k = \(\frac{1}{10}\)

∴ The probability distribution is:

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 = 1 or 6k = 1

⇒ k = \(\frac{1}{6}\).

The probability distribution is as given below:

Question 10.

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

Solution:

Let H denotes the success of getting a head.

Sample space {TTT, THH, THT, HTT, THH, HTH, HHT, HHH}

Zero success = {TTT}

Question 11.

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

Solution:

Let X denotes the number sixes.

Probability of getting one six on one dice = \(\frac{1}{6}\).

Probability of getting 1, 2, 3, 4 or 5 on a dice = \(\frac{5}{6}\).

0 success ⇒ Six does not appear on any dice.

∴ Probability of getting 1, 2, 3, 4, 5 on both dice

= \(\frac{5}{6}\) × \(\frac{5}{6}\) = \(\frac{25}{36}\).

i.e., P(0) = \(\frac{25}{36}\).

One six on one of the dice will occur as (Six, Non-six), (Non- six, Six).

∴ Probability of getting a success

P(1) = \(\frac{1}{6}\) × \(\frac{5}{6}\) + \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{10}{36}\).

Probability of getting sixes on both dice

P(2) = \(\frac{1}{6}\) × \(\frac{1}{6}\) = \(\frac{1}{36}\).

∴ Probability distribution is:

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 selected in 5 ways.

∴ No. of ways of selecting two numbers without replacement out of 6 positive integers = 6 × 5 = 30.

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:

When two dice are rolled, the number of exhaustive cases = 6 × 6 = 36.

∴ Probability distribution:

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 the mean, variance and standard deviation of X.

Solution:

There are 15 students in a class. Each has the same chance of being choosen.

The probability of each student to be selected = \(\frac{1}{15}\).

Mean = 17.53, Variance = 4.78, S.D. (X) = 2.19.

Question 15.

In a meeting, 70% of the members favojir 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:

Question 16.

The means of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is

(A) 1

(B) 2

(C) 3

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

Solution:

The variables are 1, 2 and 5.

1 is written on 3 faces.

∴ Probability of getting 1 = \(\frac{3}{6}\) = \(\frac{1}{2}\).

2 is written on two faces.

∴ Probablility of getting 2 = \(\frac{2}{6}\) = \(\frac{1}{3}\).

5 is written on one face.

∴ Probability of getting 5 = \(\frac{1}{6}\).

Probability distribution is:

∴ Part (B) is the correct answer.

Question 17.

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then, value of E(X) is

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

(B) \(\frac{27}{221}\)

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

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

Solution:

(i) When two cards are drawn, let no ace occurs.

Two non ace cards may be drawn in ^{48}C_{2} ways

= \(\frac{48×47}{2}\) = 24 × 47 = 1128.

Out of 52 cards, any two cards may be drawn in ^{52}C_{2}

= \(\frac{52×51}{2}\) = 26 × 51 = 1326 ways

∴ Probability of getting no ace = \(\frac{1128}{1326}\).

(ii) One ace and one non-ace cards may be drawn in

^{4}C_{1} × ^{48}C_{1} = 4 × 48 = 192 ways.

∴ Probability of getting one ace and one non-ace card

= \(\frac{1128}{1326}\).

(iii) The number of ways of drawing two aces = ^{4}C_{2} = 6

∴ Probability of getting 2 aces = \(\frac{6}{1326}\).

∴ Probability distribution is:

∴ Part (D) is the correct answer.