Ncert Solutions For Class 12 Maths Ex 13.5

Ncert Solutions For Class 12 Maths Chapter 13 Ex 13.5

Q.1: A die will be thrown for 6 times. Let us consider that, “getting any odd number “will be the success, then what the probability is of

(a) Getting 5 successes? (b) Getting at least 5 successes? (c) Getting at most 5 successes?

Solution:

Odd numbers on the die are 1, 3 and 5.

When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any odd numbers in the experiment of 6 toss trials.

In a single throw of the die, the probability of getting any odd number is a = \(\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\)

Thus,

b = 1 – A = \(\frac{ 1 }{ 2 }\)

Y has the binomial distribution.

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 6 }C_{ r } \frac{ 1 }{ 2 }^{ 6 – r } \frac{ 1 }{ 2 }^{ r }\)

= \(^{ 6 }C_{ r } \frac{ 1 }{ 2 }^{ 6 }\)

 

(a) Probability of getting 5 successes

P(5 successes) = P(Y = 5)

= \(^{ 6 }C_{ 5 } \frac{ 1 }{ 2 }^{ 6 }\)

= \(\frac{ 6! }{ 5! \times 1! } \frac{ 1 }{ 2 }^{ 6 }\)

= \(6 \times \frac{ 1 }{ 64 }\)

= \(\frac{ 3 }{ 32 }\)

 

(b) Probability of getting at least 5 successes

P(at least 5 successes) = P(Y ≥ 5)

= P(Y = 5) + P(Y = 6)

= \(^{ 6 }C_{ 5 } \frac{ 1 }{ 2 }^{ 6 } + ^{ 6 }C_{ 6 } \frac{ 1 }{ 2 }^{ 6 }\)

= \(\frac{ 6! }{ 5! \times 1! } \frac{ 1 }{ 2 }^{ 6 } + \frac{ 6! }{ 6! \times 0! } \frac{ 1 }{ 2 }^{ 6 } \)

= \(6 \times \frac{ 1 }{ 64 } + 1 \times \frac{ 1 }{ 64 } \)

= \(\frac{ 7 }{ 64 }\)

 

(c) Probability of getting at most 5 successes

P(at most 5 successes) = P(Y ≤ 5)

= 1 – P(Y > 5)

= 1 – P(Y = 6)

= 1 – \(^{ 6 }C_{ 6 } \frac{ 1 }{ 2 }^{ 6 }\)

= 1 – \(\frac{ 6! }{ 6! \times 0! } \frac{ 1 }{ 2 }^{ 6 }\)

= 1 – \(1 \times \frac{ 1 }{ 64 }\)

= 1 – \(\frac{ 1 }{ 64 }\) = \(\frac{ 63 }{ 64 }\)

 

Q.2: A pair of dice will be thrown for 4 times. Let us consider that, “getting a doublet” will be the success, then what is the probability for getting two successes?

Solution:

When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any doublets in the experiment of 4 toss trials.

In a single throw of the die, the probability of getting any odd number is a = \(\frac{ 6 }{ 36 } = \frac{ 1 }{ 6 }\)

Thus,

b = 1 – A = \(1 – \frac{ 1 }{ 6 } = \frac{ 6 – 1 }{ 6 } = \frac{ 5 }{ 6 }\)

Y has the binomial distribution with n = 4.

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 4 }C_{ r } \frac{ 5 }{ 6 }^{ 4 – r } \frac{ 1 }{ 6 }^{ r }\)

\(^{ 4 }C_{ r } \left (\frac{ 5^{ 4 – r }}{ 6^{ 4 } } \right)\)

Probability of getting 2 successes

P(2 successes) = P(Y = 2)

= \(^{ 4 }C_{ 2 } \left (\frac{ 5^{ 4 – 2 }}{ 6^{ 4 } } \right)\)

= \(\frac{ 4! }{ 2! \times 2! } \frac{ 5^{ 2 }}{ 1296 } \)

= \(\frac{ 4 \times 3 \times 2! }{ 2! \times 2! } \times \frac{ 25 }{ 1296 }\)

= \(\frac{ 12 }{ 2 } \times \frac{ 25 }{ 1296 }\)

= \(6 \times \frac{ 25 }{ 1296 }\) = \(\frac{ 25 }{ 216 }\)

 

Q.3: In a large bulk of different items, there are 5% of the defective items. Find the probability for the sample having 10 items, which will include at most one defective item only.

Solution:

As the drawing of items will be done with replacement, so the trial will be Bernoulli trials.

Let us consider X which denotes the total number of defective items of the sample from which 10 items are drawn successively.

Thus,

p = \(\frac{ 5 }{ 100 } = \frac{ 1 }{ 20 }\)

⟹ q = 1 – p

⟹ q = 1 – \(\frac{ 1 }{ 20 } = \frac{ 19 }{ 20 }\)

Y has the binomial distribution with n = 4 and p = \(\frac{ 1 }{ 20 }\).

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 4 }C_{ r } \frac{ 19 }{ 20 }^{ 4 – r } \frac{ 1 }{ 20 }^{ r }\)

P(at most 1 defective item) = P(X ≤ 1)

= P(X = 0) + P(X = 1)

= \(^{ 10 }C_{ 0 } \left (\frac{ 19 }{ 20 } \right)^{ 10 – 0 }. \left (\frac{ 1 }{ 20 } \right)^{ 0 } + ^{ 10 }C_{ 1 } \left (\frac{ 19 }{ 20 } \right)^{ 10 – 1 }. \left (\frac{ 1 }{ 20 } \right)^{ 1 }\)

= \(\frac{ 10! }{ 10! 0! } \left (\frac{ 19 }{ 20 } \right)^{ 10 }. \times 1 + \frac{ 10! }{ 9! 1! } \left (\frac{ 19 }{ 20 } \right)^{ 9 }. \left (\frac{ 1 }{ 20 } \right)\)

= \(1 \times \left (\frac{ 19 }{ 20 } \right)^{ 10 }. \times 1 + \frac{ 10 \times 9! }{ 9! 1! } \left (\frac{ 19 }{ 20 } \right)^{ 9 }. \left (\frac{ 1 }{ 20 } \right)\)

= \(\left (\frac{ 19 }{ 20 } \right)^{ 10 } \times 1 + 10 \times \left (\frac{ 19 }{ 20 } \right)^{ 9 }. \left (\frac{ 1 }{ 20 } \right)\)

= \(\left (\frac{ 19 }{ 20 } \right)^{ 9 } \left [ \frac{ 19 }{ 20 } + \frac{ 10 }{ 20 } \right ]\)

= \(\left (\frac{ 19 }{ 20 } \right)^{ 9 } \left (\frac{ 29 }{ 20 } \right)\) = \(\left (\frac{ 29 }{ 20 } \right). \left (\frac{ 19 }{ 20 } \right)^{ 9 }\)

 

Q.4: From a deck of 52 cards, well- shuffled, five cards will be drawn successively with having replacement. Find the probability that-

(a) All the five cards chosen will be shades?

(b) Among the five selected cards, only three of them will be spade?

(c) None of them is a spade?

Solution:

As the drawing of cards will be done with replacement, so the trial will be Bernoulli trials.

Let us consider X which denotes the number of spade cards in the deck from which 5 cards are drawn successively.

We know that in a deck of cards, there are 52 cards and among them 13 cards are spade.

Thus,

a = \(\frac{ 13 }{ 52 } = \frac{ 1 }{ 4 }\)

⟹ b = 1 – a

⟹ b = 1 – \(\frac{ 1 }{ 4 } = \frac{ 3 }{ 4 }\)

Y has the binomial distribution.

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 4 }C_{ r } \frac{ 19 }{ 20 }^{ 4 – r } \frac{ 1 }{ 20 }^{ r }\)

(a) P(all the five cards chosen will be spades) = P(Y = 5) having n = 5 and p = \(\frac{ 1 }{ 4 }\)

= \(^{ 5 }C_{ 5 } . \left (\frac{ 3 }{ 4 } \right)^{ 5 – 5 }. \left (\frac{ 1 }{ 4 } \right)^{ 5 }\)

= \(\frac{ 5! }{ 5! 0! } . \left (\frac{ 3 }{ 4 } \right)^{ 0 }. \left (\frac{ 1 }{ 4 } \right)^{ 5 }\)

= \(1 \times 1 \times \frac{ 1 }{ 1024 }\)

= \(\frac{ 1 }{ 1024 }\)

 

(b) P(among them 3 cards chosen will be spades) = P(Y = 3) having n = 5 and p = \(\frac{ 1 }{ 4 }\)

= \(^{ 5 }C_{ 3 } . \left (\frac{ 3 }{ 4 } \right)^{ 5 – 3 }. \left (\frac{ 1 }{ 4 } \right)^{ 3 }\)

= \(\frac{ 5! }{ 3! 2! } . \left (\frac{ 3 }{ 4 } \right)^{ 2 }. \left (\frac{ 1 }{ 4 } \right)^{ 3 }\)

= \(5 \times 2 \times \frac{ 9 }{ 16 } \times \frac{ 1 }{ 64 }\)

= \(\frac{ 45 }{ 512 }\)

 

(iii) P(none of them will be spades) = P(Y = 0) having n = 5 and p = \(\frac{ 1 }{ 4 }\)

= \(^{ 5 }C_{ 0 } . \left (\frac{ 3 }{ 4 } \right)^{ 5 – 0 }. \left (\frac{ 1 }{ 4 } \right)^{ 0 }\)

= \(\frac{ 5! }{ 0! 5! } . \left (\frac{ 3 }{ 4 } \right)^{ 5 } \times 1 \)

= \(1 \times \frac{ 243 }{ 1024 } \)

= \(\frac{ 243 }{ 1024 }\)

 

Q.5: If the probability that a bulb which is produced by the factory will fuse at least after 150 days of its use, is 0.10. What will be the probability that out of 5 such bulbs,

(a) None

(b) Not more than one

(c) More than 1

(d) At least 1

Bulb will fuse after 150 days of its use.

Solution:

As the drawing of bulbs will be done with replacement, so the trial will be Bernoulli trials.

Let us consider X which denotes the number of bulbs which will fuse at least after 150 days of its use in the experiment on 5 trial bulbs.

As per the data in the Q,

a = 0.1

⟹ b = 1 – 0.1 = 0.9

Y has the binomial distribution.

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

Here, n = 5 and p = 0.1.

Thus,

\(^{ 5 }C_{ r } \left(0.9 \right)^{ 5 – r } \left(0.1 \right)^{ r }\)

(a) P(none) = P(Y = 0)

= \(^{ 5 }C_{ 0 } \left(0.9 \right)^{ 5 – 0 } \left(0.1 \right)^{ 0 }\)

= \(\frac{ 5! }{ 0! 5! } \left(0.9 \right)^{ 5 } \times 1\)

= (0.9)5 = 0.59049 = 0.59

 

(b) P(not more than 1) = P(Y ≤ 1)

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

= \(^{ 5 }C_{ 0 } \left(0.9 \right)^{ 5 – 0 } \left(0.1 \right)^{ 0 } + ^{ 5 }C_{ 1 } \left(0.9 \right)^{ 5 – 1 } \left(0.1 \right)^{ 1 } \)

= \(\frac{ 5! }{ 0! 5! } \left(0.9 \right)^{ 5 } \times 1 + \frac{ 5! }{ 1! 4! } \left(0.9 \right)^{ 4 } \times 0.1 \)

= (0.9)5 + \(\frac{ 5 \times 4! }{ 1! 4! } \left(0.9 \right)^{ 4 } \times 0.1 \)

= (0.9)5 + 5 × (0.9)4 × 0.1

= 0.59049 + 0.5 × (0.9)4

= 0.59049 + 0.32805 = 0.91854

(c) P(more than 1) = P(Y ≥ 1)

= 1 – P(Y ≤ 1) = 1 – 0.91854 = 0.08146

 

(d) P(at least one) = P(Y > 1)

= 1 – P(Y < 1) = 1 – P(Y = 0) = 1 – 0.59 = 0.41

 

Q.6: There are 10 balls in the bag, each marked with any one of the digit from 0 to 9. Find the probability that none of them will be marked with the digit 0, if the four balls will be drawn successively from the bag, with replacement?

 

Solution:

As the drawing of balls will be done with replacement, so the trial will be Bernoulli trials.

Let us consider X which denotes the number of balls which will be marked with any one of the digits from 0 to 9 from which 4 balls are drawn successively from the bag.

As per the data in the Q.,

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

⟹ b = 1 – \(\frac{ 1 }{ 10 } = \frac{ 9 }{ 10}\)

Y has the binomial distribution.

Hence, P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

Here, n = 4 and p = \(\frac{ 1 }{ 10 }\).

Thus,

\(^{ 4 }C_{ r } \frac{ 9 }{ 10 }^{ 4 – r } \frac{ 1 }{ 10 }^{ r }\)

Now,

P(none of them is marked with 0) = P(Y = 0)

= \(^{ 4 }C_{ 0 }\left (\frac{ 9 }{ 10 } \right)^{ 4 – 0 }. \left (\frac{ 1 }{ 10 } \right)^{ 0 }\)

= \(\frac{ 4! }{ 0! 4! } .\left (\frac{ 9 }{ 10 } \right)^{ 4 } \times 1\)

= \(1 \times \left (\frac{ 9 }{ 10 } \right)^{ 4 } \times 1\) = \(\left (\frac{ 9 }{ 10 } \right)^{ 4 }\)

 

 

Q.7: There are 20 true and false type Q.s which are asked in the Q. paper in an examination. Let us consider a situation that a student picks up his option by tossing a fair coin every time. If its heads on the coin, the student answered true for the Q., and if its tails then the student answered false. What will be the probability that he can answer at least 12 Q.s.

Solution:

Let, Y represents the total number of correctly answered Q.s from those 20 Q.s.

As the picking of answer from the repeated tossing will be done with replacement, so the trial will be Bernoulli trials.

As per the student’s observation, whenever there is head on the coin it represents that the answer is true and whenever there is tails on the coin it represents that the answer is false.

Thus,

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

⟹ b = 1 – p = 1 – \(\frac{ 1 }{ 2 }\) = \(\frac{ 1 }{ 2 }\)

Y will have the binomial distribution with n = 20, a = \(\frac{ 1}{ 2 }\) and b = \(\frac{ 1}{ 2 }\)

Thus,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 20 }C_{ r } \left(\frac{ 1 }{ 2 } \right)^{ 20 – r } \left(\frac{ 1 }{ 2 } \right)^{ r }\)

= \(^{ 20 }C_{ r } \left(\frac{ 1 }{ 2 } \right)^{ 20 }\)

P(at least 12 Q.s from those 20 Q.s) = P(Y ≥ 12)

= P(Y = 12) + P(Y = 13) + P(Y = 14) + …….. + P(Y = 20)

= \(^{ 20 }C_{ 12 } \left(\frac{ 1 }{ 2 } \right)^{ 20 }\) + \(^{ 20 }C_{ 13 } \left(\frac{ 1 }{ 2 } \right)^{ 20 }\) + \(^{ 20 }C_{ 14 } \left(\frac{ 1 }{ 2 } \right)^{ 20 }\) + …………………. + \(^{ 20 }C_{ 20 } \left(\frac{ 1 }{ 2 } \right)^{ 20 }\)

= \(\left (\frac{ 1 }{ 2 } \right)^{ 2 }. \left [ ^{ 20 }C_{ 12 } + ^{ 20 }C_{ 13 } + ^{ 20 }C_{ 14 } + ……… + ^{ 20 }C_{ 20 } \right ]\)

 

Q.8: Let us consider Y being a binomial distribution R(6, \(\frac{ 1 }{ 2 }\)). Prove that, Y = 3 will be the most likely outcome.

Solution:

Here, as per the data given in the Q.,

Y is the binomial distribution with having binomial distribution as R(6, \(\frac{ 1 }{ 2 }\)).

Hence,

n = 6 and a = \(\frac{ 1 }{ 2 }\)

⟹ b = 1 – p = 1 – \(\frac{ 1 }{ 2 }\) = \(\frac{ 1 }{ 2 }\)

Y will have the binomial distribution with n = 20, a = \(\frac{ 1}{ 2 }\) and b = \(\frac{ 1}{ 2 }\)

Thus,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 6 }C_{ r } \left(\frac{ 1 }{ 2 } \right)^{ 6 – r } \left(\frac{ 1 }{ 2 } \right)^{ r }\)

= \(^{ 6 }C_{ r } \left(\frac{ 1 }{ 2 } \right)^{ 6 }\)

Here, we can see that P(Y = r) can be the maximum if, \(^{ 6 }C_{ r }\) can be maximum.

Thus,

\(^{ 6 }C_{ 0 } = ^{ 6 }C_{ 6 } = \frac{ 6! }{ 0! 6! } = 1\) \(^{ 6 }C_{ 1 } = ^{ 6 }C_{ 5 } = \frac{ 6! }{ 1! 5! } = \frac{ 6 \times 5! }{ 1! 5! } = 6\) \(^{ 6 }C_{ 2 } = ^{ 6 }C_{ 4 } = \frac{ 6! }{ 2! 4! } = \frac{ 6 \times 5 \times 4! }{ 2! 4! } = 3 \times 5 = 15\) \(^{ 6 }C_{ 3 } = ^{ 6 }C_{ 3 } = \frac{ 6! }{ 3! 3! } = \frac{ 6 \times 5 \times 4 \times 3! }{ 3! 3! } = 2 \times 5 \times 2 = 15\)

Here, we can see that the value of \(^{ 6 }C_{ 3 }\) is maximum.

Hence, for y = 3, P(Y = r) is maximum.

Therefore, Y = 3 is the most likely outcome.

 

 

Q.9: In a multiple choice Q.s examination which having three possible answers for every five Q.s, find the probability for the candidate that he will get 4 or more correct answers by having a guess every time.

Solution:

Let, Y represents the total number of correctly answered Q.s among those 5 multiple choice Q.s by having a guess.

As the picking of answer from the repeated guessing for the correct answer from the multiple choices will be done with replacement, so the trial will be Bernoulli trials.

Now,

Probability of getting the correct answer is given by,

a = \(\frac{ 1 }{ 3 }\)

⟹ b = 1 – a = 1 – \(\frac{ 1 }{ 3 }\) = \(\frac{ 2 }{ 3 }\)

Hence,

Y will have the binomial distribution with n = 5, a = \(\frac{ 1}{ 3 }\) and b = \(\frac{ 2 }{ 3 }\)

Thus,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 5 }C_{ r } \left(\frac{ 2 }{ 3 } \right)^{ 5 – r } \left(\frac{ 1 }{ 3 } \right)^{ r }\)

P(guessing at least 4 correct answer) = P(Y ≥ 4)

= P(Y = 4) + P(Y = 5)

= \(^{ 5 }C_{ 4 } \left(\frac{ 2 }{ 3 } \right)^{ 5 – 4 } \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \(^{ 5 }C_{ 5 } \left(\frac{ 2 }{ 3 } \right)^{ 5 – 5 } \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \(^{ 5 }C_{ 4 } \left(\frac{ 2 }{ 3 } \right)^{ 1 } \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \(^{ 5 }C_{ 5 } \left(\frac{ 2 }{ 3 } \right)^{ 0 } \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \(^{ 5 }C_{ 4 } . \frac{ 2 }{ 3 } . \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \(^{ 5 }C_{ 5 } . 1 . \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \(\frac{ 5! }{ 4! 1! } . \frac{ 2 }{ 3 } . \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \(\frac{ 5! }{ 5! 0! } . 1 . \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \(\frac{ 5 \times 4! }{ 4! } . \frac{ 2 }{ 3 } . \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \( 1 . 1 . \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \( 5 . \frac{ 2 }{ 3 } . \left(\frac{ 1 }{ 3 } \right)^{ 4 }\) + \( 1 . 1 . \left(\frac{ 1 }{ 3 } \right)^{ 5 }\)

= \(\frac{ 10 }{ 243 } + \frac{ 1 }{ 243 }\) = \(\frac{ 11 }{ 243 }\)

 

Q.10: A person purchased a lottery ticket from 50 different lotteries, among each of which his chance to win a prize is \(\frac{ 1 }{ 100 }\). Find the probability that the person will win a prize

(i) Exactly once

(ii) at least once

(iii) at least twice?

Solution:

Let us consider Y which represents the number of winning prizes among 50 lotteries.

As the picking of winner will be done with replacement, so the trial will be Bernoulli trials.

Now,

Y is a binomial distribution with n = 50 and a = \(\frac{ 1 }{ 100 }\)

⟹ b = 1 – a = 1 – \(\frac{ 1 }{ 100 }\) = \(\frac{ 99 }{ 100 }\)

Thus,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 50 }C_{ r } \left(\frac{ 99 }{ 100 } \right)^{ 50 – r } \left(\frac{ 1 }{ 100 } \right)^{ r }\)

 

(a) P (winning exactly once) = P(X = 1)

= \(^{ 50 }C_{ 1 } \left(\frac{ 99 }{ 100 } \right)^{ 50 – 1 } \left(\frac{ 1 }{ 100 } \right)^{ 1 }\)

= \(\frac{ 50! }{ 1! 49! } \left(\frac{ 99 }{ 100 } \right)^{ 49 } \left(\frac{ 1 }{ 100 } \right)^{ 1 }\)

= \(50 \left(\frac{ 1 }{ 100 } \right) \left(\frac{ 99 }{ 100 } \right)^{ 49 } \)

= \(\frac{ 1 }{ 2 } \left(\frac{ 99 }{ 100 } \right)^{ 49 } \)

 

(ii) P(winning at least once) = P(Y ≥ 1)

= 1 – P(Y < 1)

= 1 – P(X = 0)

= 1 – \(^{ 50 }C_{ r } \left(\frac{ 99 }{ 100 } \right)^{ 50 }\)

= 1 – \(\frac{ 50! }{ 0! 50! }. \left(\frac{ 99 }{ 100 } \right)^{ 50 }\)

= 1 – \(1 . \left(\frac{ 99 }{ 100 } \right)^{ 50 }\)

= 1 – \(\left(\frac{ 99 }{ 100 } \right)^{ 50 }\)

 

(iii) P (at least twice) = P(Y ≥ 2)

= 1 – P(Y < 2)

= 1 – P(Y ≤ 1)

= 1 – [P(Y = 0) + P(Y = 1)]

= [1 – P(Y = 0)] – P(Y = 1)

= 1 – \( \left(\frac{ 99 }{ 100 } \right)^{ 49 } \). \(\left [ \frac{ 99 }{ 100 } + \frac{ 1 }{ 2 } \right ]\)

= 1 – \( \left(\frac{ 99 }{ 100 } \right)^{ 49 } \). \(\left (\frac{ 99 + 50 }{ 100 } \right)\)

= 1 – \(\left (\frac{ 99 + 50 }{ 100 } \right)\) . \( \left(\frac{ 99 }{ 100 } \right)^{ 49 } \)

 

 

Q.11: What will be the probability of getting 5 exactly twice in 7 throws of a die?

Solution:

As the repeatedly tossing of a die will be done with replacement, so the trial will be Bernoulli trials.

Let us consider Y which represents the number of times to get 5 in 7 throws of the die.

As per the data given in the Q., a = \(\frac{ 1 }{ 6 }\)

⟹ b = 1 – a = 1 – \(\frac{ 1 }{ 6 }\) = \(\frac{ 5 }{ 6 }\)

Y is the probability distribution with n = 7, a = \(\frac{ 1 }{ 6 }\) and b = \(\frac{ 5 }{ 6 }\)

Then,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 7 }C_{ r } \left(\frac{ 5 }{ 6 } \right)^{ 7 – r } \left(\frac{ 1 }{ 6 } \right)^{ r }\)

 

P (getting 5 exactly twice) = P (Y = 2)

= \(^{ 7 }C_{ 2 } \left(\frac{ 5 }{ 6 } \right)^{ 7 – 2 } \left(\frac{ 1 }{ 6 } \right)^{ 2 }\)

= \(\frac{ 7! }{ 2! 5! } \left(\frac{ 5 }{ 6 } \right)^{ 2 } \left(\frac{ 1 }{ 6 } \right)^{ 2 }\)

= \(\frac{ 7 \times 6 \times 5! }{ 2 \times 5! } \left(\frac{ 5 }{ 6 } \right)^{ 2 } \left(\frac{ 1 }{ 6 } \right)^{ 2 }\)

= \(\frac{ 7 \times 6 }{ 2 } \left(\frac{ 5 }{ 6 } \right)^{ 2 } \left(\frac{ 1 }{ 36 } \right) \)

= \(21 \left(\frac{ 5 }{ 6 } \right)^{ 2 } \left(\frac{ 1 }{ 36 } \right) \)

= \( \left(\frac{ 5 }{ 6 } \right)^{ 2 } \left(\frac{ 7 }{ 12 } \right) \)

 

 

Q.12: 10% of the certain articles which are manufactured will be defective. Find the probability for the samples having 12 such articles among which 9 of them will be defective.

 

Solution:

As the repeated selection of the articles in a random sample space will be done with replacement, so the trial will be Bernoulli trials.

Let us consider Y which represents the number of times to select the defective articles in the random sample space of 12 articles.

As per the data given in the Q., a = 10% = \(\frac{ 10 }{ 100 } = \frac{ 1 }{ 10 }\)

b = 1 – a = 1 – \(\frac{ 1 }{ 10 }\) = \(\frac{ 9 }{ 10 }\)

Y is the probability distribution with n = 12, a = \(\frac{ 1 }{ 10 }\) and b = \(\frac{ 9 }{ 10 }\)

Then,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 12 }C_{ r } \left(\frac{ 9 }{ 10 } \right)^{ 12 – r } \left(\frac{ 1 }{ 10 } \right)^{ r }\)

P(selecting 9 defective articles) = \(^{ 12 }C_{ 9 } \left(\frac{ 9 }{ 10 } \right)^{ 12 – 9 } \left(\frac{ 1 }{ 10 } \right)^{ 9 }\)

= \(\frac{ 12! }{ 9! 3! } \left(\frac{ 9 }{ 10 } \right)^{ 3 } \left(\frac{ 1 }{ 10 } \right)^{ 9 }\)

= \(\frac{ 12 \times 11 \times 10 \times 9! }{ 9! \times 3 \times 2 \times 1 } \left(\frac{ 9 }{ 10 } \right)^{ 3 } \left(\frac{ 1 }{ 10 } \right)^{ 9 }\)

= \(\frac{ 12 \times 11 \times 10 }{ 3 \times 2 \times 1 } \left(\frac{ 9 }{ 10 } \right)^{ 3 } \left(\frac{ 1 }{ 10 } \right)^{ 9 }\)

= \(220 . \left(\frac{ 9 }{ 10 } \right)^{ 3 } \left(\frac{ 1 }{ 10 } \right)^{ 9 }\)

= \(\frac{ 22 \times 9^{ 3 } }{ 10^{ 11 } }\)

 

 

Q.13: Among the 100 bulbs in the box, 10 bulbs are defective. The probability that out of the sample of 5 bulbs, none of the bulbs are defective is,

(a) 10-1

(b) \(\left (\frac{ 1 }{ 2 } \right)^{ 5 }\)

(c) \(\left (\frac{ 9 }{ 10 } \right)\)

(d) \(\left (\frac{ 9 }{ 10 } \right)^{ 5 }\)

 

Solution:

As the repeated selection of the defective bulbs in the box will be done with replacement, so the trial will be Bernoulli trials.

Let us consider Y which represents number of the defective bulbs in a random sample of 5 bulbs.

As per the data given in the Q., a = \(\frac{ 10 }{ 100 } = \frac{ 1 }{ 10 }\)

⟹ b = 1 – a = 1 – \(\frac{ 1 }{ 10 }\) = \(\frac{ 9 }{ 10 }\)

Y is the probability distribution with n = 5, a = \(\frac{ 1 }{ 10 }\) and b = \(\frac{ 9 }{ 10 }\)

Then,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 5 }C_{ r } \left(\frac{ 9 }{ 10 } \right)^{ 5 – r } \left(\frac{ 1 }{ 10 } \right)^{ r }\)

P(none of the bulbs are defective) = P(Y = 0)

\(^{ 5 }C_{ 0 } \left(\frac{ 9 }{ 10 } \right)^{ 5 – 0 } \left(\frac{ 1 }{ 10 } \right)^{ 0 }\)

= \(\frac{ 5! }{ 5! 0! } \left(\frac{ 9 }{ 10 } \right)^{ 5 } \times 1\)

= \(1 \times \left(\frac{ 9 }{ 10 } \right)^{ 5 } \times 1\)

= \(\left(\frac{ 9 }{ 10 } \right)^{ 5 }\)

Hence, the correct option is (d).

 

Q.14: The probability that the student will not be a swimmer is \(\frac{ 1 }{ 5 }\). The probability that from a group of 5 students, 4 of them will be a swimmer is

(a) \( \left (\frac{ 4 }{ 5 } \right)^{ 4 }\frac{ 1 }{ 5 }\)

(b) \(^{ 5 }C_{ 4 }\left (\frac{ 4 }{ 5 } \right)^{ 4 }\frac{ 1 }{ 5 }\)

(c) \(^{ 5 }C_{ 1 }\left (\frac{ 4 }{ 5 } \right)^{ 4 }\frac{ 1 }{ 5 }\)

(d) None of these

Solution:

As the repeated selection of the students in the group who all are swimmers will be done with replacement, so the trial will be Bernoulli trials.

Let us consider Y which represents number of the students from the group of 5 students, who all are swimmers.

As per the data given in the Q., b = \(\frac{ 1 }{ 5 }\)

⟹ a = 1 – b = 1 – \(\frac{ 1 }{ 5 }\) = \(\frac{ 4 }{ 5 }\)

Y is the probability distribution with n = 5, a = \(\frac{ 4 }{ 5 }\) and b = \(\frac{ 1 }{ 5 }\)

Then,

P(Y = r) = \(^{ n }C_{ r } b^{ n – r } a^{ r }\), where n = 0, 1, 2, 3, …………, n

= \(^{ 5 }C_{ r } \left(\frac{ 1 }{ 5 } \right)^{ 5 – r } \left(\frac{ 4 }{ 5 } \right)^{ r }\)

P(none of the bulbs are defective) = P(Y = 4)

= \(^{ 5 }C_{ 4 } \left(\frac{ 1 }{ 5 } \right)^{ 5 – 4 } \left(\frac{ 4 }{ 5 } \right)^{ 4 }\)

= \(^{ 5 }C_{ 4 } \left(\frac{ 1 }{ 5 } \right) \left(\frac{ 4 }{ 5 } \right)^{ 4 }\)

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