Chapter 13 Probability Ex 13.5 Solutions
Question - 1 : - A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) At least 5 successes?
(iii) At most 5 successes?
Answer - 1 : -
We know that therepeated tosses of a dice are known as Bernoulli trials.
Let the number ofsuccesses of getting an odd number in an experiment of 6 trials be x.
Probability of gettingan odd number in a single throw of a dice (p)
Thus, q = 1 – p = ½
Now, here x has abinomial distribution.
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2 …n
= 6Cx (1/2)6-x (1/2)x
= 6Cx (1/2)6
(i) Probability ofgetting 5 successes = P(X = 5)
= 6C5 (1/2)6
= 6 ×1/64
= 3/32
(ii) Probability ofgetting at least 5 successes = P(X ≥ 5)
= P(X = 5) + P(X = 6)
= 6C5 (1/2)6 + 6C5 (1/2)6
= 6 ×1/64 + 6 ×1/64
= 6/64 + 1/64
= 7/64
(iii) Probability ofgetting at most 5 successes = P(X ≤ 5)
We can also write itas: 1 – P(X>5)
= 1 – P(X = 6)
= 1 – 6C6 (1/2)6
= 1 – 1/64
= 63/64
Question - 2 : - A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer - 2 : -
We know that therepeated tosses of a pair of dice are known as Bernoulli trials.
Let the number oftimes of getting doublets in an experiment of throwing two dice simultaneouslyfour times be x.
Thus, q = 1 – p = 1 –1/6 = 5/6
Now, here x has abinomial distribution, where n = 4, p = 1/6, q = 5/6
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2, … n
= 4Cx (5/6)4-x (1/6)x
= 4Cx (54-x/66)
Hence, Probability ofgetting 2 successes = P(X = 2)
= 4C2 (54-2/64)
= 6 (52/64)
= 6 × (25/1296)
= 25/216
Question - 3 : - There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Answer - 3 : -
Let there be x numberof defective items in a sample of ten items drawn successively.
Now, as we can seethat the drawing of the items is done with replacement. Thus, the trials areBernoulli trials.
Now, probability ofgetting a defective item, p = 5/100 = 1/20
Thus, q = 1 – 1/20 =19/20
∴ We can say thatx has a binomial distribution, where n = 10 and p = 1/20
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2 …n
Probability of gettingnot more than one defective item = P(X ≤1)
= P(X = 0) + P(X = 1)
= 10C0 (19/20)10(1/20)0 +10C1 (19/20)9(1/20)1
Question - 4 : - Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) All the five cards are spades?
(ii) Only 3 cards are spades?
(iii) None is a spade?
Answer - 4 : -
Let the number ofspade cards among the five drawn cards be x.
As we can observe thatthe drawing of cards is with replacement, thus, the trials will be Bernoullitrials.
Now, we know that in adeck of 52 cards there are total 13 spade cards.
Thus, Probability ofdrawing a spade from a deck of 52 cards
= 13/52 = ¼
q = 1 – ¼ = 3/4
Thus, x has a binomialdistribution with n = 5 and p = ¼
Thus, P(X = x) = nCx qn-x px ,where x = 0, 1, 2, …n
Question - 5 : - The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use.
Answer - 5 : -
Let us assume that thenumber of bulbs that will fuse after 150 days of use in an experiment of 5trials be x.
As we can see that thetrial is made with replacement, thus, the trials will be Bernoulli trials.
It is alreadymentioned in the question that, p = 0.05
Thus, q = 1 – p = 1 –0.05 = 0.95
Here, we can clearlyobserve that x has a binomial representation with n = 5 and p = 0.05
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2… n
= 5Cx (0.95)5-x(0.05)x
(i) Probability of nosuch bulb in a random drawing of 5 bulbs = P(X = 0)
= 5C0 (0.95)5-0(0.05)0
= 1× 0.955
= (0.95)5
(ii) Probability ofnot more than one such bulb in a random drawing of 5 bulbs = P (X≤ 1)
= P(X = 0) + P(X = 1)
= 5C0 (0.95)5-0(0.05)0+ 5C1(0.95)5-1(0.05)1
= 1× 0.955 +5 × (0.95)4 × 0.05
= (0.95)4 (0.95+0.25)
= (0.95)4 ×1.2
(iii) Probability ofmore than one such bulb in a random drawing of 5 bulbs = P (X>1)
= 1 – P(X ≤ 1)
= 1 – [(0.95)4 ×1.2]
(iv) Probability of atleast one such bulb in a random drawing of 5 bulbs = P (X ≥ 1)
= 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – (0.95)5
Question - 6 : - A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Answer - 6 : -
Let us assume thatnumber of balls with digit marked as zero among the experiment of 4 balls drawnsimultaneously be x.
As we can see that theballs are drawn with replacement, thus, the trial is a Bernoulli trial.
Probability of a balldrawn from the bag to be marked as digit 0 = 1/10
It can be clearlyobserved that X has a binomial distribution with n = 4 and p = 1/10
Thus, q = 1 – p = 1 –1/10 = 9/10
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2, …n
Question - 7 : - In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.
Answer - 7 : -
Let us assume that thenumber of correctly answered questions out of twenty questions be x.
Since, ‘head’ on thecoin shows the true answer and the ‘tail’ on the coin shows the false answers.Thus, the repeated tosses or the correctly answered questions are Bernoullitrails.
Thus, p = ½ and q = 1– p = 1 – ½ = ½
Here, it can beclearly observed that x has binomial distribution, where n = 20 and p = ½
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2 … n
Question - 8 : - Suppose X has a binomial distribution B (6, ½) . Show that X = 3 is the most likely outcome.
(Hint: P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6)
Answer - 8 : -
Given X is any randomvariable whose binomial distribution is B (6, ½)
Thus, n = 6 and p = ½
q = 1 – p = 1 – ½ = ½
Thus, P(X = x) = nCx qn-x px,where x = 0, 1, 2 …n
It can be clearlyobserved that P(X = x) will be maximum if 6cx willbe maximum.
∴6cx = 6c6 =1
6c1 = 6c5 =6
6c2 = 6c4 =15
6c3 = 20
Hence we can clearlysee that 6c3 is maximum.
∴ for x = 3, P(X = x)is maximum.
Hence, proved that themost likely outcome is x = 3.
Question - 9 : - On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer - 9 : -
In this question, wehave the repeated correct answer guessing form the given multiple choicequestions are Bernoulli trials
Let us now assume, Xrepresents the number of correct answers by guessing in the multiple choice set
Now, probability ofgetting a correct answer, p = 1/3
Thus, q = 1 – p = 1 –1/3 = 2/3
Clearly, we have X isa binomial distribution where n = 5 and P = 1/3
Question - 10 : - A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100. What is the probability that he will win a prize
(a) At least once
(b) Exactly once
(c) At least twice?
Answer - 10 : -
(a) Let X representsthe number of prizes winning in 50 lotteries and the trials are Bernoullitrials
Here clearly, we haveX is a binomial distribution where n = 50 and p = 1/100
Thus, q = 1 – p = 1 –1/100 = 99/100
