# ISI2017-MMA-27

840 views

A box contains $5$ fair and $5$ biased coins. Each biased coin has a probability of head $\frac{4}{5}$. A coin is drawn at random from the box and tossed. Then the second coin is drawn at random from the box ( without replacing the first one). Given that the first coin has shown head, the conditional probability that the second coin is fair is

1. $\frac{20}{39}\\$
2. $\frac{20}{37}\\$
3. $\frac{1}{2}\\$
4. $\frac{7}{13}$

edited

prob of head from fair coin= (5*1/2)/((5*1/2)+(5*4/5))=5/13

prob of head from biased coin= (5*4/5)/((5*1/2)+(5*4/5))=8/13

now probability of second coin to be fair is given as =  (head from baised coin*probability of fair coin if biased coin was selected at first )+(head from fair coin*probability of fair coin if fair coin was selected at first)= 8/13*5/9 + 5/13*4/9 = 20/39.

Ans-A

selected by
0
5/13 * 4/9 is 20/117
0
i don't know what is was thinking.. edited. P(A):The probability of getting head is $\frac{5}{10}*\frac{1}{2}+\frac{5}{10}*\frac{4}{5}=\frac{13}{20}$

Probability that second coin is= first fair * second fair + first unfair * second fair=

P(B):Probability that second coin is fair= $\frac{1}{4}*\frac{4}{9}+\frac{2}{5}*\frac{5}{9}=\frac{1}{3}$

conditional probability = $\frac{P(B)}{P(A)}=\frac{\frac{1}{3}}{\frac{13}{20}}=\frac{20}{39}$

Option A
0
Sir, Can you please explain it using a diagram?

I'm having difficulty in understanding it.
0
Sir, can you please explain the P(B) part in detail? I am not getting the value of P(B)
1

1/4*4/9 here we are taking the case that first coin was fair and now choosing second, so for second draw we have to choose 1 out of 4 fair coin and coin be any fair out of 9

1/5 * 5/9 here we are taking case were first coin was unfair, rest logic is same

@rajatmyname

0
Sir, for the first coin why the probability is not 5/10 * 4/9? I am not getting 1/4 term you used for first case
0

@rajatmyname let me be more clear 1*1/4*4/9 here

1 - First coin is head and fair

1/4 - since the first coin we consider as fair, we have now 4 fair coins left so selecting 1 out of 4

4/9 - that one fair coin can be any out of total 9 coins so 4/9

0
@tesla so probability of first unfair and second fair should be 1/5 * 5/9. Is it correct ?
0
Probability of first fair will be 5/10

second fair ( first fair) 4/9

Second fair (first unfair) 5/9

Based on that either 5/10 * 4/9 or 5/10*5/9
0

@Tesla! Sir

how the probability of first unfair is 2/5 ...mention the above solution which u gave

plz explain...i am not understand

0
@hardesh patel

first unfair is 2/5 because it means the first coin is unfair plus we have given after first draw we get head so in case of unfair coin the probability of getting head is 4/5 and prob of getting unfair coin is 5/10 and both events are independent so we finally get first unfair coin given it shows head is 5/10(for unfair)*4/5(for getting head in unfair coin)=2/5.Hope it will help you

i tried  to give ans more clearly, hope it is understandable. ## Related questions

1 vote
1
218 views
Let $X_1$, and $X_2$ and $X_3$ be chosen independently from the set $\{0, 1, 2, 3, 4\}$, each value being equally likely. What is the probability that the arithmetic mean of $X_1, X_2$ and $X_3$ is the same as their geometric mean? $\frac{1}{5^2}\\$ $\frac{1}{5^3}\\$ $\frac{3!}{5^3}\\$ $\frac{3}{5^3}$
Suppose that $X$ is chosen uniformly from $\{1, 2, \dots , 100\}$ and given $X=x, \: Y$ is chosen uniformly from $\{1, 2, \dots , x\}$. Then $P(Y =30)=$ $\frac{1}{100}$ $\frac{1}{100} \times (\frac{1}{30}+ \dots + \frac{1}{100})$ $\frac{1}{30}$ $\frac{1}{100} \times (\frac{1}{1}+ \dots +\frac{1}{30})$
There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. The probability that only two tests are required is $\frac{1}{2}\\$ $\frac{1}{3}\\$ $\frac{1}{4}\\$ $\frac{1}{6}$
There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. The probability that only two tests are required is $\left(\dfrac{1}{2}\right)$ $\left(\dfrac{1}{3}\right)$ $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{6}\right)$