Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(d) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(d)

Question 1.

State which of the following is the probability distribution of a random variable X with reasons to your answer:

(a)

X = x | 0 | 1 | 2 | 3 | 4 |

p(x) | 0.1 | 0.2 | 0.3 | 0.4 | 0.1 |

(b)

X = x | 0 | 1 | 2 | 3 |

p(x) | 0.15 | 0.35 | 0.25 | 0.2 |

(c)

X = x | 0 | 1 | 2 | 3 | 4 | 5 |

p(x) | 0.4 | R | 0.6 | R^{2} |
0.7 | 0.3 |

Solution:

(a) As ∑p_{i} = 1.1 > 1, the given distribution is not a probability distribution.

(b) As ∑p_{i} = 0.95 < 1, the given distribution is not a probability distribution.

(c) As the value of R is not known, the given distribution is not a probability distribution.

Question 2.

Find the probability distribution of number of doublets in four throws of a pair of dice. Find also the mean and the variance of the number of doublets.

Solution:

Let the random variable X represents the number of doublets in 4 throws of a pair of dice.

X can take values 0, 1, 2, 3, 4

In a single throw of two dice

X can take values 0, 1, 2, 3, 4

In a single throw of two dice.

P(doublet) = \(\frac{6}{36}\)

P(non-doublet) = 1 – P (doublet) = \(\frac{5}{36}\)

Clearly the given experiment is a binomial experiment with n = 4,

p = \(\frac{1}{6}\), q = \(\frac{5}{6}\)

N.B. We can use the definition to find mean and variance.

Question 3.

Four cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. Calculate the mean and variance of the number of aces.

Solution:

Let the random variable X denotes the number of aces.

Thus X can take values 0, 1, 2, 3, 4.

Clearly the given experiment is a binomial experiment with n = 4

N.B. We can use the definition to find mean and variance.

Question 4.

Find the probability distribution of

(a) number of heads in three tosses of a coin.

Solution:

Let the random variable X denotes the number of heads in three tosses of a coin.

X can take values 0, 1, 2, 3

In one toss p(H) = \(\frac{1}{2}\), p(T) = \(\frac{1}{2}\)

This experiment is a binomial experiment with n = 3, p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

(b) number of heads in simultaneous tosses of four coins.

Solution:

When 4 coins tossed simultaneously, let the random variable X denotes the number of heads.

X can take values 0, 1. 2, 3, 4

This experiment is a binomial experiment with n = 4, p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

Question 5.

A biased coin where the head is twice as likely to occur as the tail is, tossed thrice. Find the probability distribution of number of heads.

Solution:

Let in a toss P(T) = x

According to the question P(H) = 2x

Now 2x + x = 1

⇒ x = \(\frac{1}{3}\)

Thus P(T) = \(\frac{1}{3}\), P(H) = \(\frac{3}{3}\)

Clearly the given experiment is a binomial experiment with n = 3, p = P(H) = \(\frac{2}{3}\), q = P(T) = \(\frac{1}{3}\)

Let the random variable X denotes the number of heads in three throws of that coin.

X can take values 0, 1, 2, 3

Question 6.

Find the probability distribution of the number of aces in question no. 3 if the cards are drawn successively without replacement.

Solution:

Total number of cards = 52

Number of aces = 4

Number of cards drawn = 4 (one by one without replacement)

Let X = the random variable of number of aces drawn.

X can take values 0, 1, 2, 3 or 4

Question 7.

From a box containing 32 bulbs out of which 8 are defective 4 bulbs are drawn at random successively one after another with replacement. Find the probability distribution of the number of defective bulbs.

Solution:

Total number of bulbs = 32

Number of defective bulbs = 8

∴ Number of nondefective bulbs = 24

Number of bulbs drawn = 4 (one after another with replacement)

Clearly the experiment is a binomial experiment with n = 4

p = p ((drawing a defective bulb) = \(\frac{8}{32}\) = \(\frac{1}{4}\)

q = p (drawing a non defective bulb) = \(\frac{3}{4}\)

Let the random variable X denotes the number of defective bulbs.

∴ X can take values 0, 1, 2, 3, 4.

Question 8.

A random variable X has the following probability distribution.

X = x | 0 | 1 | 2 | 3 | 4 | 5 |

p(x) | 0 | R | 2R | 3R | 3R | R |

Determine

(a) R

(b) P(X < 4)

(d) P(2 ≤ X ≤ 5)

(c) P(X ≥ 2)

Solution:

(a) Clearly ∑P_{i} = 1

⇒ R + 2R + 3R + 3R + R = 1

Question 9.

Find the mean and the variance of the number obtained on a throw of an unbiased coin.

Solution:

When an unbiased coin is tossed we donot get any number.

p(getting a number) = 0

Thus mean = variance = 0

If instead of coin it would be an unbiased die

Then let X = The number obtained in the throw.

X can take values 1, 2, 3, 4, 5 or 6.

P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4)

Question 10.

A pair of coins is tossed 7 times. Find the probability of getting

(i) exactly five tails

(ii) at least five tails

(iii) at most five tails

Solution:

Clearly the given experiment is a binomial experiment with

Question 11.

If a pair of dice is thrown 5 times then find the probability of getting three doublets.

Solution:

In a single through of a pair of dice

p (a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

p (a non doublet) = 1 – p (a doublet) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

Clearly the given experiment is a binomial experiment with n = 5

p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)

p(3 doublets in 5 throw) = ^{5}C_{3} p^{3}q^{2 }

= 10. \(\left(\frac{1}{6}\right)^3\) \(\left(\frac{5}{6}\right)^2\) = \(\frac{250}{6^5}\) = \(\frac{125}{3888}\)

Question 12.

Four cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that:

(i) all the four cards are diamonds

(ii) only two cards are diamonds

(iii) none of the cards is a diamond.

Solution:

Out of 52 cards there are 13 diamonds. 4 cards are drawn one by one with replacement.

When we draw a card

Question 13.

In an examination, there are twenty multiple-choice questions each of which is supplied with four possible answers. What is the probability that a candidate would score 80% or more in the answers to these questions?

Solution:

Total number of questions = 20 for each question

p (correct answer) = \(\frac{1}{4}\)

p (wrong answer) = \(\frac{3}{4}\)

p (the score is > 80%)

= p (no. of correct answer > 16)

= p (16 correct answers)

+ p(17 correct answers)

+ p (18 correct answers)

+ p (19 correct answers)

+ p (20 correct answers)

Question 14.

A bag contains 7 balls of different colours. If five balls are drawn successively with replacement then what is the probability that none of the balls drawn is white?

Solution:

Total number of balls = 7 (The colours are different)

Number of balls drawn = 5 (one by one with replacement)

Case – 1

(If a white coloured ball is not present)

p(non is white) = 1

Case – 2

(If one ball is white).

In one draw p(white ball) = \(\frac{1}{7}\)

p (non white ball) = \(\frac{6}{7}\)

When 5 balls are drawn

p (none of the balls drawn is white) = ^{5}C_{0 }\(\left(\frac{1}{7}\right)^0\) \(\left(\frac{6}{5}\right)^5\) = \(\left(\frac{6}{7}\right)^5\)

Question 15.

Find the probability ofthrowing at least 3 sixes in 5 throws of a die.

Solution:

In one throw of a die

Question 16.

The probability that a student securing first division in an examination is \(\). What is the probability that out of 100 students twenty pass in first division?

Solution:

Clearly the given experiment is a binomial experiment with

Question 17.

Sita and Gita throw a die alternatively till one of them gets a 6 to win the game. Find their respective probability of winning if Sita starts first.

Solution:

In one throw of a die

p (getting a 6) = \(\frac{1}{6}\)

p (not getting a 6) = \(\frac{5}{6}\)

If Sita begins the game she may win in first round or second round or third round, etc.

Thus p(Sita wins)

Question 18.

If a random variable X has a binomial distribution B(8, \(\frac{1}{2}\)) then find X for which the outcome is the most likely. [Hint: Find X=x for which P(X = x) is the maximum, x = 0, 1, 2, 3,…….. 8.]

Solution:

Given binomial distribution is B(8, \(\frac{1}{2}\))

Thus n = 8, p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)

We shall find X which is most likely i.e. for which p(x = r) is maximum where r = 0, 1, 2, ….., 8

As p = q, the probability is maximum when ^{8}Cr is maximum

But ^{8}Cr, r = 0, 1, 2, ….. ,8 is maximum when r = 4.

Thus the most likely outcome is x = 4.