ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.8

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.8

Question 1.
A coin is tossed 5 times. What is the probability of getting
(i) 3 heads ?
(ii) atmost 3 heads ?
Answer:
p = prob. of getting a head = \(\frac{1}{2}\)
∴ q = 1 – p
= 1 – \(\frac{1}{2}=\frac{1}{2}\)
and n = 5
Since the events are independent so it is a case of binomial distribution.
Then P(X = r) = ⁿC_r p^r
= ⁵C_r\(\left(\frac{1}{2}\right)^{5-r}\left(\frac{1}{2}\right)^r\)

(i) P(X = 3) = ⁵C₃\(\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3\)
= ⁵C₃\(\left(\frac{1}{2}\right)^5=\frac{5 \times 4}{2} \times \frac{1}{32}=\frac{5}{16}\)

(ii) P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
= ⁵C₀\(\left(\frac{1}{2}\right)^5\) + ⁵C₁\(\left(\frac{1}{2}\right)^5\) + ⁵C₂\(\left(\frac{1}{2}\right)^5\) + ⁵C₃\(\left(\frac{1}{2}\right)^5\)
= [1 + 5 + \(\frac{5 \times 4}{2}+\frac{5 \times 4}{2}\)]\(\left(\frac{1}{2}\right)^5=\frac{26}{32}=\frac{13}{16}\)

Question 1(Old).
Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell ^ whether or not the trials of drawing balls are Bernoullian trials when after each draw the ^ ball drawn is
(i) replaced ‘
(ii) not replaced in the urn. (NCERT)
Answer:
(i) Total no. of bulbs = 7 + 9 = 16
p = probability of drawing a one ball from urn = \(\frac{1}{16}\)
Since ball is replaced every time.
So prob. of success p remains same for every trial
i. e. (p) does not change from trial to trial, in this case, trials of drawing balls are Bemoullian trials.

(ii) When after each drawing ball is not replaced.
Here trials are not independent i.e. result of any trial is not independent of other trial. Since probability of success (p) changes from trial to trial so in this case, trials of drawing balls are not Bemoullions trials.

Question 2.
A coin is tossed 5 times. Find the probability of getting
(i) atleast 4 heads
(ii) atmost 4 heads.
Answer:
Since coin is tossed 5 times so events are independent so given problem is a case of binomial distribution.
Let p = prob. of getting head in a single throw of coin = \(\frac{1}{2}\)
and q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\) and n = 5
Then by binomial distribution, we have
P(X = r) = ⁿC_rp^rq^n-r
= ⁵C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}\)
= ⁵C_r\(\left(\frac{1}{2}\right)^5\)

(i) P (X ≥ 4) = P (X = 4) + P (X = 5)
= ⁵C₄\(\left(\frac{1}{2}\right)^5\) + ⁵C₅\(\left(\frac{1}{2}\right)^5\)
= (5 + 1)\(\frac{1}{2^5}\)
= \(\frac{6}{32}=\frac{3}{16}\)

(ii) P(X ≤ 4) = 1 – P(X = 5)
= 1- ⁵C₅\(\left(\frac{1}{2}\right)^5\)
= 1 – \(\frac{1}{32}=\frac{31}{32}\)

Question 2(Old).
Find out whether each of the following is Bemoullian trial or not knowing that 10 trials of each experiment are held :
(i) tosses of a coin.
(ii) drawing balls with replacement from a bag containing 5 white balls only.
(iii) throws of a pair of dice.
(iv) attempting 10 true-false type of questions on the basis of outcomes of tosses of a coin.
Answer:
(i) While tossing of coin,
p = prob. of success = \(\frac{1}{2}\)
q = prob. of failure = \(\frac{1}{2}\)
In each trial, p does not change from trial to trial. So result of any trial is independent of other trial. So given trial be a Bemoullian trial.

(ii) In this trial, balls are drawn with replacement. So result of any trial is independent of other trial. So p (prob. of success) does not change from trial to trial. So given trial be a Bemoullian trial.

(iii) Given trial of throwing a pair of dice is a bemoullian trial. Since result of any trial is independent of other trial and probability of success p does not change from trial to trial.

(iv) Since probability of getting head or tail in each toss of a unbiased coin is \(\frac{1}{2}\). So result of any trial is independent of other trial and p does not changes from trial to trial.
Thus given trial be a bemoullian trial.

Question 3.
A fair coin is tossed 6 times. What are the parameters of this binomial experiment ? What is the probability of getting exactly 4 heads ? Atleast 4 heads ?
Answer:
When a fair coin is tossed.
Then S = {H, T}
p = prob. of success
= prob. of getting head = \(\frac{1}{2}\)
Here n = 6, q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\)
Thus given distribution is a binomial distribution with parameters n = 6 and P = \(\frac{1}{2}\)
Also P (r) = ⁿC_r p^r q^n-r
= ⁶C_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{6-r}\)
⇒ P(r) = ⁶C_r\(\left(\frac{1}{2}\right)^6\)

required probability = P (X = 4)
= ⁶C₄\(\left(\frac{1}{2}\right)^6=\frac{6 \times 5}{2} \times \frac{1}{2^6}=\frac{30}{128}=\frac{15}{64}\)

Prob. of getting atleast 4 heads = P (X ≥ 4)
= P (X = 4) + P (X = 5) + P (X = 6)
= \({ }^6 \mathrm{C}_4\left(\frac{1}{2}\right)^6+{ }^6 \mathrm{C}_5\left(\frac{1}{2}\right)^6+{ }^6 \mathrm{C}_6\left(\frac{1}{2}\right)^6\)
= \(\left[\frac{6 \times 5}{2}+6+1\right] \frac{1}{2^6}=\frac{22}{2^6}=\frac{11}{32}\)

Question 4.
If X denotes the number of heads in a single toss of 4 fair coins, then find
(i) P (X = 3)
(ii) P (X < 2)
(iii) P (X ≤ 2)
(iv) P (1 < X ≤ 3).
Answer:
When a coin is tossed
Then p = prob. of getting head = \(\frac{1}{2}\);
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\)
Since a coin is tossed 4 times or single toss of 4 coins, events are independent. it is a problem of binomial distribution with
n = 4 and p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
also X ~ B (4, \(\frac{1}{2}\))

P(X = r) = ⁿC_r p^r q^n-r

Question 5.
Ten coins are tossed. What is the probability of getting atleast 8 heads ? (NCERT Exemplar)
Answer:
Ten coins are tossed simultaneously is same as tossing of one coin 10 times. So there are 10 trials and all the trials are independent. So it is a problem of binomial distribution.
p = prob. of getting head = \(\frac{1}{2}\);
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\)

So we have a binomial distribution with
n = 10, p = q = \(\frac{1}{2}\)
Also P(r) = ⁿC_r p^r q^n-r
= ¹⁰C_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{10-r}\)
= ¹⁰C_r \(\left(\frac{1}{2}\right)^{10}\)

Probability of getting atleast 8 heads
= P(X ≥ 8)
= P(X = 8) + P(X = 9) + P(X = 10)

Question 6.
Four cards are drawn one by one with replacement from a well-shuffled deck of playing cards. Find the probability that atleast three cards are of diamond.
Answer:
Since the cards are drawn one by one with replacement so events are independent. So it is a problem of binomial distribution
p = prob. of drawing a diamond card

Question 7.
A coin is tossed four times. If X is the number of heads observed, then find the ‘probability distribution of X. (NCERT)
Answer:
Since a coin is tossed four times so we have 4 trials and all trials are independent. So it is a problem of binomial distribution.
Random variable X denotes the no. of heads observed.
p = prob. of getting head = \(\frac{1}{2}\) ;
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\); n = 4
Now P (X = r) = ⁿC_r p^r q^n-r
= ⁴C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{4-r}\)

∴ P(X = r) = ⁴C_r\(\left(\frac{1}{2}\right)^4\)
and X takes values 0, 1,2, 3, 4.

Thus, required probability distribution of X is given below:

Question 8.
Find the probability distribution of the number of tails when five fair coins are ?? tossed simultaneously.
Answer:
Since five coins are tossed simultanenously is same as the tossing of a coin 5 times. So we have 5 trials and all trials are independent. So we have a problem of binomial distribution.
p = prob. of getting trial = \(\frac{1}{2}\)
q = 1 – \(\frac{1}{2}=\frac{1}{2}\); n = 5
∴ P (r) = ⁿC_r p^r q^n-r
= ⁵C_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}\)
= ⁵C_r \(\left(\frac{1}{2}\right)^5\)

Since X be the random variable denotes the no. of tails observed.
∴ X take values 0, 1, 2, 3, 4, 5.

Thus, required probability distribution of X is given below :

Question 9.
Find the probability distribution of the number of successes in two tosses of a die when a success is defined as getting a number greater than 4. (NCERT)
Answer:
p = probability of success i.e. getting a number greater than 4 {i. e. 5 and 6} = \(\frac{2}{6}=\frac{1}{3}\)
Then q = 1 – p = 1 – \(\frac{1}{3}=\frac{2}{3}\); n = 2
Let X denotes the number of successes in 2 tosses of a die.
Then by binomial distribution, we have ,

Then the probability distribution of X is given below :

Question 10.
A pair of dice is tossed twice. If the random variable X is defined as the ^0 number of doublets, find the probability distribution of X.
Answer:
A pair of dice tossed twice so we have 2 trials and both are independent. So it is a problem of binomial distribution.
When a pair of dice tossed we have 36 equally likely events, p = prob. of getting a doublet = \(\)

Since favourable outcomes are
{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 1)}
q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\); n = 2

Since X be the random variable defined as the number of doublets and X can takes values 0, 1,2.

The required probability distribution of X is given below :

Question 11.
Find the probability distribution of number of doublets in three throws of a pair of dice. (NCERT)
Answer:
When a pair of dice are thrown, the total no. of outcomes are 36 and all outcomes are equally likely, out of these 36 oucomes, 6 outcomes are doublets i.e.
{(1, 1), (2, 2), (3, 3) (4, 4), (5, 5), (6, 5)}
p = prob. of getting a doublet = \(\frac{6}{36}=\frac{1}{6}\)
q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\) and n = 3

Let the random variable X denotes the no. of doublets
since n = 3, X can take values 0, 1, 2, 3.

The probability distribution of X is given below:

Question 12.
From a well-shuffled pack of 52 cards, 3 cards are drawn one by one with replacement. Find the probability distribution of number of queens.
Answer:
Since cards are drawn one by one with replacement.
∴ the events are independent. So it is a problem of binomial distribution.

p = prob. of getting a queen = \(\frac{4}{52}=\frac{1}{13}\)
q = 1 – p = 1 – \(\frac{4}{52}=\frac{1}{13}\)
Here n = 3
Let X be the binomial variate denotes the no. of queens.
X can takes values 0, 1, 2, 3.
The probability distribution function of X is given by

Thus, the probability distribution of X is given below:

Question 13.
A pair of dice is thrown 3 times. If getting a total of 10 is considered a success, then find the probability distribution of the number of successes.
Answer:
When a pair of dice are thrown, total no. of possible outcomes is equal to 36 and all are equally likely.
When a pair of dice thrown 3 times, we have 3 trials and all are independent. So it is a problem of binomial distribution.
Here p = probability of getting 10 = \(\frac{3}{36}=\frac{1}{12}\)
q = 1 – p = 1 – \(\frac{1}{12}=\frac{11}{12}\)
Since favourable outcomes are
{(4, 6), (5, 5),.(6, 4)}
Here n = 3. Let X denotes binomial variate and getting a sum of 10 in 3 draws of pair of dice.
X can takes values 0, 1, 2, 3

Thus, the required probability distribution of X is given below :

Question14.
An urn contains 5 white and 3 blue balls. If two balls are drawn one by one with replacement, then find the probability distribution of the number of white balls drawn.
Answer:
Since balls are drawn one by one with replacement
the events are independent hence it is a problem of binomial distribution.
Total no. of balls in an urn = 5 + 3 = 8

p = probability of drawing a white ball = \(\frac{5}{8}\)
q = 1 – p = 1 – \(\frac{5}{8}=\frac{3}{8}\) Here n = 2

Let X be the binomial variate denotes the no. of red balls drawn in 2 draws.
and X can takes values 0, 1, 2 3.

Thus, the probability distribution of X is given below

Question 15.
An urn contains 4 white and 3 red balls. Find the probability distribution of the x> number of red balls when 3 balls are drawn one by one with replacement.
Answer:
Since balls are drawn one by one with replacement.
∴ the events are independent. So it is a problem of binomial distribution.
Here, Total number of balls in an urn = 4 + 3 = 7
∴ p = probability of drawing a red ball = \(\frac{3}{7}\)
∴ q = 1 -p = 1 – \(\)Here n = 2
Let X be the random variable denotes the no. of white balls drawn in 3 draws.
∴ X can takes values 0, 1, 2, 3.

Thus, the probability distribution of X is given below :

Question 16.
Five bad eggs are mixed with 10 good ones. If three eggs are drawn one by one with replacement, then find the probability distribution of the number of good eggs drawn. (ISC 2010)
Answer:
No. of bad eggs = 5
No. of good eggs = 10
∴ Total no. of eggs = 5+10=15
Since the eggs are drawn one by one with replacement so it is a problem of binomial distribution.
p = prob. of drawing a good egg = \(\frac{10}{15}=\frac{2}{3}\)
and q = 1 – p = 1 – \(\frac{10}{15}=\frac{2}{3}\) and n = 3
Let X denote the no. of good eggs drawn in three draws of good eggs then X can take values 0,1,2,3.
Thus by binomial distribution, we have

Question 16(Old).
(i) Four dice are thrown simultaneously. If the occurrence of an odd number in a single dice is considered a success, find the probability of atmost 2 successes. (ISC 2006)
(ii) Five dice are thrown simultaneously. If the occurrence of an odd number is considered a success, find the probability of maximum three successes. (ISC 2015)
Answer:
(i) Since throwing of four dice simultaneously is same as the throw of single dice 4 times.
Here we have 4 trials and all are independent.
Hence it is a problem of binomial distribution.
When a dice is thrown, S = {1, 2, 3, 4, 5, 6} and all outcomes are equally likely.
p = prob. of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)
Since favourable outcomes are {1, 3, 5}
∴ q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\) Here n = 4

Thus P(X = r) = ⁿC_r p^r q^n-r
= ⁴C_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{4-r}\)

Probability of atmost 2 success = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)

(ii) Since throwing of five dice simultaneously is same as the throw of a single dice 5 times.
Here we have 5 trials and all are independent.
When a dice is throw, possible outcomes are {1, 2, 3, 4, 5, 6} and all are equally likely.
∴ p = prob. of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)
and q = 1 – p = 1 – \(\frac{3}{6}=\frac{1}{2}\) and Hence n = 5

So we have a binomial distribution with n = 5, p = q = \(\frac{1}{2}\)
P(r) = ⁿC_r p^r q^n-r
= ⁵C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}\)
= ⁵C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}\)

∴ required prob. of maximum three success = P (X ≤ 3) = 1 – P (X = 4) – P (X = 5)

Question 17.
From a lot of 30 bulbs which include 6 defective bulbs, a sample of 4 bulbs is drawn at ^ random with replacement Find the probability distribution of the number of defective ^ bulbs drawn. (NCERT)
Answer:
Let p = probability of defective bulb = \(\frac{6}{30}=\frac{1}{5}\)
q = 1 – p = 1 – \(\frac{1}{5}=\frac{4}{5}\); n = 4
Let X denotes the number of defective bulbs drawn in a sample of 4 bulbs drawn successively with replacement.
Then by binomial distribution, we have

Therefore the probability distribution of X is given below :

Question 17(Old).
The probability that a student entering a university will graduate is 0.4. Find the probability that out of 3 students of the university
(i) none will be graduate
(ii) only one will be graduate
(iii) all will be graduate.
Answer:
p be the probability that a student will graduate is 0-4.
Then q = 1 -p = 1 – 0 4 = 0.6 ; n = 3
Let X denotes the number of students will graduate in a random sample of 3 students
Then by binomial distribution, we have P(X = r) = ³C_r (0.4)^r (0.6)^3-r; r = 0, 1, 2, 3
(i) required probability that none will graduate
= P (X = 0)
= ³C₀(0.4)⁰ (0.6)³
= 0.216

(ii) required probability that only one will graduate = P(X = 1)
= ³C₁(0.4)¹ (0.6)²
= 3 × 0.4 × 0.36
= 0.432

(iii) required probability that all will graduate = P (X = 3)
= ³C₃(0.4)³ (0.6)⁰
= 0.064

Question 18.
A die is thrown 5 times. Find the pro* bability that an odd number will turn up
(i) exactly 3 times
(ii) atleast 4 times
(iii) maximum 3 times. (ISC 2015)
Answer:
Since a dice is thrown one by one so events are independent. So it is a problem of binomial distribution.
p = prob. of getting an odd number turn up = \(\frac{3}{6}=\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\) and n = 5
Thus by binomial distribution, we have

Question 19.
An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.
Answer:
Let p = probability of success
and q = prob. of failure
∴ P = 2q
Since p + q = 1
⇒ 3q = 1
⇒ q = \(\frac{1}{3}\)
P = \(\frac{2}{3}\); n = 6
Let X denotes the number of successes in 6 trials.
Then by binomial distribution, we have
P (X = r) = ⁶C_r\(\left(\frac{2}{3}\right)^r\left(\frac{1}{3}\right)^{6-r}\); r = 0,1, 2,…,6
Thus required prob. of atlest 4 successes
= P (X ≥ 4)
= P (X = 4) + P (X = 5) + P (X = 6)

Question 20.
A pair of dice is thrown 6 times. Getting a total of 7 on the two dice is considered a success. Find the probability of getting
(i) atleast 5 successes
(ii) exactly 5 successes.
Answer:
Here n = 6;p = probability of getting 7
Since favourable cases are {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

Question 21.
A pair of dice is thrown 4 times. If ‘getting a doublet’ is considered a success, 9* then find the probability of atleast 2 successes. (NCERT)
Answer:
Let p be the probability of getting doublet in a single throw of a pair of dice.
Here, favourable cases
{(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
then p = \(\frac{6}{36}=\frac{1}{6}\) ∴ q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)
Here, n = 4
Let X denotes the number of doublets in 4 throws of pair of dice.
Then by binomial distribution, we have
P(X = r) = ⁿC_rp^rq^n-r = ⁿC_r\(\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{4-r}\)
reqd. prob. of getting atfeast two successes
= P(X ≥ 2)
= 1 – P(X = 0) – P(X = 1)

Question 22.
A die is thrown 4 times. Getting a ‘1 or 6’ is considered a success. Find the probability of getting
(i) exactly 3 successes
(ii) exactly 4 successes
(iii) atmost 2 successes.
Answer:
When a die thrown, the possible outcomes are {1, 2, 3, 4, 5, 6}. All the six outcomes are equally likely. Here we have 4 trials and all trials are independent. So it is a problem of binomial distribution.
p = probability of getting 1 or 6 = \(\frac{2}{6}=\frac{1}{3}\)
q = 1 – p = 1 – \(\frac{2}{6}=\frac{1}{3}\)
Here n = 4, Now P (X = r) = ⁿC_rp^r q^n-r
= ⁴C_r\(\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{4-r}\)
(i) Required probability of getting exactly 3 successes
= P(X = 3) = ⁴C₃\(\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)\)
= 4 × \(\frac{1}{27} \times \frac{2}{3}=\frac{8}{81}\)

(ii) Required probability of getting exactly 4 successes
= P(X = 4) = ⁴C₄\(\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^0=\frac{1}{3^4}=\frac{1}{81}\)

(iii) Required probability of getting atmost 2 successes
= P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
= \({ }^4 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^4+{ }^4 C_1 \frac{1}{3}\left(\frac{2}{3}\right)^3+{ }^4 C_2\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2\)
= \(\frac{16}{81}+\frac{4 \times 8}{81}+\frac{6 \times 4}{3^4}\)
= \(\frac{16+32+24}{81}=\frac{72}{81}=\frac{8}{9}\)

Question 23.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective ? (NCERT)
Answer:
p = prob. of deflective item = \(\frac{10}{100}=\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}=\frac{9}{10}\); n = 2

We know that, by binomial distribution
P(r) = ¹²C_r\(\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{12-r}\)
required prob. = P (9) = ¹²C₉\(\)
= \(\frac{12 \times 11 \times 10}{6} \times \frac{9^3}{10^{12}}\)
= 220 x \(\frac{9^3}{10^{12}}\)

Question 24.
A box contains 100 bulbs, 10 are defective. If 5 bulbs are drawn successively with replacement, find the probability that :
(i) none is defective (NCERT)
(ii) exactly two are defective.
Answer:
p = probability of getting defective bulb
= \(\frac{10}{100}=\frac{1}{10}\)
q = 1 – p
= 1 – \(\frac{1}{10}=\frac{9}{10}\)

(i) Since 5 balls are drawn successively with replacement reqd. probability
∴ reqd. probability
= q × q × q × q × q
= \(\left(\frac{9}{10}\right)^5\)

(ii) ∴ required prob. of getting exactly two defective bulbs and 3 are good ones
= pp{qqq}
= \(\left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^3\)

Question 25.
Ten eggs are drawn successively, with replacement, from a basket containing 10% defective eggs. Find the probability that there is atleast one defective egg. (NCERT)
Answer:
Let p be the probability of getting a
defective egg = \(\frac{10}{100}\)
p = \(\frac{1}{100}\) Then q = 1 – \(\frac{1}{10}=\frac{9}{10}\); n = 10
Let X denotes the no. of defective eggs in a sample of 10 eggs drawn with replacement from a lot.
Then by binomial distribution ; we have
P(X = r) = ¹⁰C_r\(\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{10-r}\) Or r = 0, 1, …, 10
required probability that there is atleast one defective egg = P (X ≥ 1)
= 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀\(\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{10}\)
= 1 – \(\left(\frac{9}{10}\right)^{10}\)

Question 26.
Find the probability of getting 5 exactly yjo twice in 7 throws of a die. (NCERT)
Answer:
Here n = 7,
p = probability of getting ‘5’ = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)
Thus by binomial distribution, we have
P(r) = ⁿC_r p^r q^n-r
= \(\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{7-r}\)

required probability = P(2)
= \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5\)
= 21 × \(\frac{5^5}{6^7}\)

Question 26.
A bag contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, then what is the probability that
(i) none is white
(ii) all are whites
(iii) only 2 are whites ?
Answer:
Since balls are drawn one by one with replacement.
So events are independent.
∴ it is a problem of binomial distribution.
Here No. of white balls = 5
No. of red balls = 7
No. of black balls = 8
∴ Total no. of balls = 5 + 7 + 8 = 20
Here p = prob. of getting white ball
= \(\frac{5}{20}=\frac{1}{4}\)

∴ q = 1 – P = 1 – \(\frac{1}{4}=\frac{3}{4}\) and here n = 4
Now P (X = r) = ⁿC_r p^r q^n-r
= ⁴C_r\(\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{4-r}\)

(i) required probability that none is white
= P(X = 0) = ⁴C₀\(\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^4=\frac{81}{256}\)

(ii) probability of getting all are white = P (X = 4)
= ⁴C₄\(\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^0=\frac{1}{256}\)

(iii) probability of getting only 2 are white
= P(X = 2) = ⁴C₂\(\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^2\)
= \(\frac{6 \times 9}{4^4}=\frac{54}{256}=\frac{27}{128}\)

Question 27.
A box contains 100 tickets, each bearing a different number from 1 to 100. If 5 tickets are drawn successively with replacement from the box, then find the probability that all tickets bear numbers divisible by 10.
Answer:
Let p be the probability of getting a ticket bear a number divisible by 10
Here favourable case are
{10, 20, 30, 40, 50, 60, 70, 80, 90, 100}
Then p = \(\frac{10}{100}=\frac{1}{10}\)
q = 1 – p = 1 – \(\frac{1}{10}=\frac{9}{10}\)
Here n = 5
Let X denotes the number of tickets bearing a number divisible by 10 in 5 draws.

Then by binomial distribution, we have
P(X = r) = ⁵C_r\(\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r}\)
r = 0, 1, 2, ….. 5
∴ required probability = P(X = 5)
= ⁵C_r\(\left(\frac{1}{10}\right)^5\left(\frac{9}{10}\right)^0=\left(\frac{1}{10}\right)^5\)

Question 29.
If the chance that any of the five telephone lines is busy at an instant is e, 0-01, then what is the probability that all lines are busy ? What is the probability that more than 3 lines are busy ?
Answer:
Since the probability of any of the five telephone lines is busy at an instant is 0.01
i.e. prob.of each event be same.
So all the five trials are independent.
∴ it is a case of binomial distribution.

Here p = probability that one of the telephonic line is busy at an instant = 0.01
q = 1 – p = 1 – 0.01 = 0.99
Here n = 5
P (r) = ⁿC_rp^r q^n-r
= ⁵C_r (0.01)^r(0.99)^5-r
required probability that all lines are busy
P (X = 5) = ⁵C₅ (0.01)⁵ (0.99)⁰
= (0.01)⁵
= \(\left(\frac{1}{100}\right)^5\)

required probability that more than 3 lines are busy
= P (X > 3) = P (X = 4) + P (X = 5)
= ⁵C₄ (0.01)4 (0.99)¹ + ⁵C₅ (0.01)5 (0.99)⁰
= \(\frac{5}{(100)^4} \frac{99}{100}+\frac{1}{(100)^5}\)
= \(\frac{495+1}{(100)^5}\)
= 496 \(\left(\frac{1}{100}\right)^5\).

Question 30.
The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting atleast twice ? (NCERT Exemplar)
Answer:
p = probability of a man hitting the target = \(\frac{1}{4}\)
q = 1 – n = 1 – \(\frac{1}{4}=\frac{3}{4}\) and n = 7
Let X denotes the number of times the man hitting the target in 7 trails. Then by binomial distribution, we have
P(X = r) = ⁷C_r

required probability of his hitting the target atleast twice
= P (X ≥ 2) = 1 – [P (X < 2)]
= 1 – [P(X = 0) + P(X = 1)]

Question 31.
During a war, 1 in 10 ships was sunk in the average in making a certain voyage. What is the probability that atleast 3 out of a convoy of 6 ships would arrive safely ?
Answer:
Given probability that the ship was sunk in making a certain voyage = \(\frac{1}{10}\)
Let p = probability that the ship arrive safely = 1 – \(\frac{1}{10}=\frac{9}{10}\)
q = 1 – p = 1 – \(\frac{9}{10}=\frac{1}{10}\) = prob. that ship was sunk.
Here n = 6
Then by binomial distribution, we have
P(X = r) = ⁿC_r p^r q^{n – r} = ⁶C_r \(\left(\frac{9}{10}\right)^r\left(\frac{1}{10}\right)^{6-r}\)
required probability = P (X ≥ 3)
= P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

Question 32.
A student answers 20 true-false types of questions by tossing a fair coin. If the coin fall heads, he answers ‘true’ and if it falls tails he answers ‘false’. Find the probability that he answers atleast 12 questions correctly. (NCERT)
Answer:
Let p = probability of getting a true answer = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 20
Let X denotes the number of correct answers given by student out of 20 questions.
Then by binomial distribution, we have
P(X = r) = ²⁰C_r\(\)
r = 0, 1, ……….., 20

required probability that he answers atleast 12 questions correctly = P (X ≥ 12)
= P (X = 12) + P (X = 13) + P (X = 14) + P (X = 15)+ P (X = 16) + P (X = 17) + P (X = 18) + P (X = 19) + P (X = 20)

Question 33.
The probability of a man hitting a target is 0.25. How many times must he fire so that the the probability of his hitting the target atleast once is greater than \(\frac{2}{3}\) ?
Answer:
Here, given probability of a man hitting a target = p = 0.25 = \(\frac{1}{4}\)
q = 1 – p = 1 – 0.25 = 0.75
Let the number of times the man fires the target be n such that probability of his hitting the target atleast once be greater than \(\frac{2}{3}\) .
i.e. P(X ≥ 1) > \(\frac{2}{3}\)
⇒ 1 – P (X = 0) > \(\frac{2}{3}\) …(1)
Then by Binomial distribution, we have

from (1); we have
Now eqn. (2) is satisfied if n is atleast 4.
Hence the man must fire atleast 4 times.

Question 33(Old).
The probability of India winning a test match against West-Indices is Find the probability that in a test match series India’s second win occurs at the third test.
Answer:
Let p – prob. of winning of india a test match = \(\frac{1}{2}\)
q = prob. of failure = 1 – p = 1 –\(\frac{1}{2}\) = \(\frac{1}{2}\)
As the second win occurs at the third test .’. required probability = P (prob. of winning exactly one win in first 2 matches) × P (prob. of winning in 3rd match)
[∵ P (X = r) = ⁿC_r p^r q^n-r]
= ²C₁ pq . p
= 2 × \(\left(\frac{1}{2}\right)^2 \times \frac{1}{2}=\frac{1}{4}\)

Question 34.
A fair die is tossed eight times. Find the probability that a third six is observed on the eighth throw.
Answer:
Let p = probability of getting a six in a single throw of die = \(\frac{1}{6}\)
q = 1 – p = 1 –\(\frac{1}{6}\) = \(\frac{5}{6}\)
As the third six is observed on the eighth throw.
required probability = P (prob. of getting exactly two sixes in first 7(throws) P(prob. of getting six in eighth throw)
= ⁷C₂ p² q⁵ . p [By binomial distribution P(X = r)]
= ⁿC_r p^r q^n-r
= \(\frac{7 \times 6}{2} \times\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}\)
= \(\frac{21}{216}\left(\frac{5}{6}\right)^3=\frac{7}{72}\left(\frac{5}{6}\right)^3\)