Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert Solutions Probability Maths
Exercise 13.5
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?
Sol: The repeated throwing of dice are Bernoulli trials. Let X denote no. of successes (getting an odd number) in an experiment of 6 trials.
Clearly, X has the binomial distribution with n =6 and p=\(\frac{1}{2}\)
\(\therefore\) P(X=x) = \(^nC_xq^{n-x}p^x\) , x = 0,1,2,...n
Here n = 6, p= \(\frac{1}{2}\), q = 1-p = \(\frac{1}{2}\),
\(\therefore\) P(X=x) = \(^6C_x(\frac{1}{2})^{n-x}(\frac{1}{2})^x= ^6C_x(\frac{1}{2})^6\)
(i) P(X=5) = \(^6C_5(\frac{1}{2})^6 = \frac{6!}{5!\times1!}\frac{1}{2^6} = \frac{3}{32}\)
(ii) P(at least 5 successes) = P(X\(\geq\)5)
= P(X=5) + P(X=6) = \(^6C_5(\frac{1}{2})^6 + ^6C_6(\frac{1}{2})^6 = \frac{6!}{5!\times1!}\frac{1}{2^6}+ \frac{6!}{6!}\frac{1}{2^6}=\frac{3}{32} +\frac{1}{64}=\frac{7}{64} \)
(iii) P(at most 5 successes) = P(X\(\leq5\)
= P(X=0) + P(X=1) + P(X=2) +P(X=3) + P(X=4) + P(X=5)
=\((\frac{1}{2})^6 + ^6C_1(\frac{1}{2})^6 + ^6C_2(\frac{1}{2})^6 + ^6C_3(\frac{1}{2})^6 + ^6C_4(\frac{1}{2})^6 + ^6C_5(\frac{1}{2})^6\)
= (1+6+15+20+15 +6)\(\frac{1}{64}\) = \(\frac{63}{64}\)
2. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Sol : Throwing of dice can be considered Bernoulli trials. Let X denote the number of successes in an experiment of n trials.
Total no of outcomes in throwing a pair of dice = 6 X 6 = 36
Doublet means same number appearing on both die in a single throw.
No of such possible outcomes is 6 which are(1,1),(2,2), (3,3), (4,4), (5,5), (6,6)
Probability of getting a doublet is (p) = \(\frac{6}{36}=\frac{1}{36}\)
Probability of not getting a doublet q = 1-p = \(\frac{5}{6}\)
X has a binomial distribution with n = 4 and p =\(\frac{1}{6}\)
P(X=x) = \(^nC_xq^{n-x}p^x=^4C_xq^{4-x}p^x=^4C_x(\frac{5}{6})^{4-x}(\frac{1}{6})^x=\)
P(X=2) =\(^4C_2(\frac{5}{6})^{4-2}(\frac{1}{6})^2=\frac{4!}{2!2!}(\frac{5}{6})^{2}(\frac{1}{6})^2=\frac{25}{216}\)
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?
Sol: Let X be the no of defective item picked in an experiment of picking 10 items(i.e 10 trials). So n=10
Probability of picking a defective item p is = 5% = \(\frac{5}{100}=\frac{1}{20}\)
Probability of not picking a defective item is = 1-p = 1-\(\frac{1}{20}=\frac{19}{20}\)
Now, P(X=x) = \(^nC_xq^{n-x}p^x=^{10}C_xq^{10-x}p^x=^{10}C_x(\frac{19}{20})^{10-x}(\frac{1}{20})^x=\)
Probability of getting not more than one defective item P(X\(\leq\)1)= P(X=0) + P(X=1)
= \(^{10}C_0(\frac{19}{20})^{10}(\frac{1}{20})^0 + ^{10}C_1(\frac{19}{20})^{10-1}(\frac{1}{20})^1\)
= \((\frac{19}{20})^{10} + 10(\frac{19}{20})^{9}(\frac{1}{20})^1\)
= \((\frac{19}{20})^9 (\frac{19}{20}+\frac{10}{20})\)
=\((\frac{19}{20})^9 (\frac{29}{20})\)
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?
Sol: Probability of getting a spade p = \(\frac{13}{52}=\frac{1}{4}\)
Probability of not getting a spade p = \(1-\frac{1}{4}= \frac{3}{4}\)
(i) Here n = 5 , x = 5
Probability of getting all the five cards as spade = P(X= 5) = \(^5C_5\left(\frac{3}{4}\right)^{5-5}\left(\frac{1}{4}\right)^5=\frac{45}{512}\)
(ii) Here n = 5, x =3
Probability of getting all the five cards as spade = P(X= 5) = \(^5C_3\left(\frac{3}{4}\right)^{5-3}\left(\frac{1}{4}\right)^3= \left(\frac{5!}{3!2!}\right)\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^3=\left(\frac{1}{4}\right)^5=\frac{1}{1024}\)
(iii) Here n = 5, x =0
Probability of getting all the five cards as spade = P(X= 5) = \(^5C_0\left(\frac{3}{4}\right)^{5-0}\left(\frac{1}{4}\right)^0= \left(\frac{3}{4}\right)^{5}=\frac{243}{1024}\)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
Sol: Probability that the bulb produced in a factory will fuse after 150 days is p = 0.05
Probability that the bulb will not fused after 150 days of use is q = 1-p = 1 - 0.05 = 0.95
(i) Here n =5, x=0
probability that none will fuse P(X=0) = \(^5C_0\left(0.95\right)^{5-0}\left(0.05\right)^0=(0.95)^5\)
(ii) Probability that not more than one bulb will fuse is P(X\(\leq0\)) = P(X=0) + P(X=1)
=\(^5C_0\left(0.95\right)^{5-0}\left(0.05\right)^0 + ^5C_1\left(0.95\right)^{5-1}\left(0.05\right)^1=(0.95)^5+5(0.05)^1(0.95)^4 = (0.95)^4(0.95+0.25)=(0.95)^4\times1.2\)
(iii) Probability that more than one will fuse is P(X>1) = 1 - Probability of not more than one
= 1-\((0.95)^4\times1.2\)
(iv) Probability that at least one bulb will fuse = 1 - probability that none will fuse = \(1-(0.95)^5\)
6. A bag consist 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?
Sol: Probability that the drawn ball is marked 0 p= \(\frac{1}{10}\)
Probability that the drwan ball is not marked 0 q = 1 -p = 1 - \(\frac{1}{10}\) =\(\frac{9}{10}\)
4 balls are drawn sonumber of trials n= 4
Let X denote the number of balls marked with 0. Hence, x=0
probability that none is marked with 0 P(x=0) = \(^4C_0\left(\frac{9}{10}\right)^{4-0}\left(\frac{1}{10}\right)^0=\left(\frac{9}{10}\right)^4\)
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 answer "false". Find the probability that he answers at least 12 questions correctly.
Sol : Here no. of trials n = 20,
probability that answer will be correct p = \(\frac{1}{2}\)
probability that answer will be false q =1- \(\frac{1}{2}\) =\(\frac{1}{2}\)
\(\therefore\) P(X=x) = \(^nC_xq^{n-x}p^x\) , x = 0,1,2,...n
p= \(\frac{1}{2}\), q = 1-p = \(\frac{1}{2}\),
\(\therefore\) P(X=x) = \(^{20}C_x(\frac{1}{2})^{n-x}(\frac{1}{2})^x= ^{20}C_x(\frac{1}{2})^{10}\)
Probability that at least 12 answers are correct P(X\(\geq\)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)
= \(^{20}C_{12}(\frac{1}{2})^{10} + ^{20}C_{13}(\frac{1}{2})^{10} + ....... + ^{20}C_{20}(\frac{1}{2})^{10}\)
= \(\left(\frac{1}{2}\right)^{10}\left[^{20}C_{12} + ^{20}C_{13}+.....+ ^{20}C_{20} \right]\)
8. Suppose X has a binomial distribution B(6,\(\frac{1}{2}\)). Show that X = 3 is the most likely outcome.
Sol: X has the binomial distribution with n =6 and p=\(\frac{1}{2}\)
\(\therefore\) P(X=x) = \(^nC_xq^{n-x}p^x\) , x = 0,1,2,...n
Here n = 6, p= \(\frac{1}{2}\), q = 1-p = \(\frac{1}{2}\),
\(\therefore\) P(X=x) = \(^6C_x(\frac{1}{2})^{n-x}(\frac{1}{2})^x= ^6C_x(\frac{1}{2})^6\)
P(X=0) = \(^6C_0(\frac{1}{2})^6 = \frac{1}{2^6} = \frac{1}{64}\)
P(X=1) = \(^6C_1(\frac{1}{2})^6 = \frac{6!}{1!\times5!}\frac{1}{2^6} = \frac{6}{64}\)
P(X=2) = \(^6C_2(\frac{1}{2})^6 = \frac{6!}{2!\times4!}\frac{1}{2^6} = \frac{15}{64}\)
P(X=2) = \(^6C_2(\frac{1}{2})^6 = \frac{6!}{2!\times4!}\frac{1}{2^6} = \frac{15}{64}\)
P(X=3) = \(^6C_3(\frac{1}{2})^6 = \frac{6!}{3!\times3!}\frac{1}{2^6} = \frac{20}{64}\)
P(X=4) = \(^6C_4(\frac{1}{2})^6 = \frac{6!}{4!\times2!}\frac{1}{2^6} = \frac{15}{64}\)
P(X=6) = \(^6C_6(\frac{1}{2})^6 = 1\times\frac{1}{2^6} = \frac{1}{64}\)
P(X=3) is maximum among all the values of X. So it is the most likely outcome.
Sol: A question has 3 options out of which one is correct.
Probability of getting the answer correct is p = \(\frac{1}{3}\)
Probability of getting the answer wrong q = 1-p =\(\frac{2}{3}\)
Here n = 5
Probability of getting 4 or more answers correct P(X\(\geq\)5) = P(X=4) + P(X=5)
= \(^5C_4\left(\frac{2}{3}\right)^{5-4}\left(\frac{1}{3}\right)^4 + ^5C_5\left(\frac{2}{3}\right)^{5-5}\left(\frac{1}{3}\right)^5=\left(5\times\frac{2}{3^5}\right)+ \frac{1}{3^5}= \frac{10}{243}+\frac{1}{243}=\frac{11}{243}\)
10. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is\(\frac{1}{100}\). What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice?
Sol: Here n = 50
Probability of winning a prize in each lottery p = \(\frac{1}{100}\)
probability of not winninga prize in each lottery q= 1 - \(\frac{1}{100}\) = \(\frac{99}{100}\)
(a) Probability of winning a prize atleast once = 1 - probability of not winning a prize at all
= 1 -P(X=0)
= 1 - \(^{50}C_0 q^{50-0} p^0\)
== 1 - \(\left(\frac{99}{100}\right)^{50}\)
(b) Probability that the person wins prize exactly once out of 50 = P(X=1)
= \(^{50}C_1 q^{50-1} p^1\)
=\(50 \left(\frac{99}{100}\right)^{50-0} \left(\frac{1}{100}\right)^1\)
= \(\left(\frac{1}{2}\right)\left(\frac{99}{100}\right)^49\)
(c)
probability of winning a prize not more than once = P(X=0) + P(X=1) = \(^{50}C_0 q^{50-0} p^0\) +\(^{50}C_1 q^{50-1} p^1\)
= \(\left(\frac{99}{100}\right)^{50}\)+\(50 \left(\frac{99}{100}\right)^{50-1} \left(\frac{1}{100}\right)^1\)
= \(\left(\frac{99}{100}\right)^{50}\)+\(\left(\frac{1}{2}\right) \left(\frac{99}{100}\right)^{50-1}\)
= \(\left(\frac{99}{100}\right)^{49}\left(\frac{99}{100}+\frac{1}{2}\right)\)
=\(\left(\frac{149}{100}\right)\left(\frac{99}{100}\right)^{49}\)
Probability of winning a prize at least twice = 1- probability of winning a prize not more than once
= 1 - \(\left(\frac{149}{100}\right)\left(\frac{99}{100}\right)^{49}\)
11. Find the probability of getting 5 exactly twice in 7 throws of a die.
Sol: Here n = 7 ,
probability of getting 5 (p) = \(\frac{1}{6}\)
probability of not getting 5(q) = 1-p = \(\frac{5}{6}\)
X= 2
Probability of getting % exactly twice is = P(X=2)
= \(^{7}C_2 q^{7-2} p^2\)
= \(\frac{7!}{2!5!} \left(\frac{5}{6}\right)^{5} \left(\frac{1}{6}\right)^2\)
= \(\frac{7.6}{2.1} \left(\frac{5}{6}\right)^{5} \left(\frac{1}{6}\right)^2\)
= \(\frac{7}{12} \left(\frac{5}{6}\right)^{5}\)
12. Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Sol: Here n = 6 ,
probability of getting 6 (p) = \(\frac{1}{6}\)
probability of not getting 6(q) = 1-p = \(\frac{5}{6}\)
Probability of getting % exactly twice is = P(X=0) + P(X=1) + P(X=2)
= \(^{6}C_0 q^{6-0} p^0\)
= \(^{6}C_0 \left(\frac{5}{6}\right)^{6} \)
= \( \left(\frac{5}{6}\right)^{6} \)
= \(\left(\frac{5}{6}\right)^4\left[\frac{25}{36}+\frac{5}{6} + \frac{15}{36}\right]\)
=\(\left(\frac{5}{6}\right)^4\left[\frac{25+30+15}{36}\right]\)
=\(\left(\frac{5}{6}\right)^4\left(\frac{70}{36}\right)\)
=\(\left(\frac{5}{6}\right)^4\left(\frac{35}{18}\right)\)
13. It is known that10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective ?
Sol : Here n =12 ,
probability that a certain article is defective is = 10% =\(\frac{10}{100}=\frac{1}{10}\)
Probability that out of 12, 9 articles are defective = P(x=9)
=
=
=\(\frac{12.11.10}{3.2}\left(\frac{9}{10}\right)^{3} \left(\frac{1}{10}\right)^9\)
= \(\left(\frac{22\times9^3}{10^2}\right) \left(\frac{1}{10}\right)^9\)
= \(\left(\frac{22\times9^3}{10^{11}}\right)\)
14. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(A) \(10^{-1}\) (B) \(\left(\frac{1}{2}\right)^5\) (C)\(\left(\frac{9}{10}\right)^5\) (D) \(\frac{9}{10}\)
Sol : probability of the bulb to be defective is(p) \(\frac{10}{100}=\frac{1}{10}\)
probability of the bulb is not defective is (q)=1 - \(\frac{1}{10}\)= \(\frac{9}{10}\)
here n = 5 , X=0 So probability that none is defective = \(^{5}C_0 q^{5-0} p^0 = q^{5}= \left(\frac{9}{10}\right)^5\)
The correct answer is option C
15. The probability that a student is not aswimmer is \(\frac{1}{5}\). Then the probability that out of five students, four are swimmer is
(A)\(^{5}C_4 \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^4\) (B)\( \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^4\)
(C)\( ^{5}C_1 \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^4\) (D) none of these
= 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)
Here n = 5, x =4
therefoe, probability that four are swimmers = \(^{5}C_4 q^{5-4} p^4=^{5}C_4 \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^4\)
The corect answer is (A)
Class 12 Ncert Solutions Probability Maths
Previous Exercies of this Chapter
Exercise 13.1 Exercixe 13.2
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Chapter 11 Three dimensional geometry
Chapter 3 Matrices
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