One Biased Coin

Recall that Bayes' Theorem states the following:

Given events $A$ and $B$ , we have that $P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$ .

Let $A$ be the event that the coin is fair, and let $B$ be the event that each of the $5$ flips results in heads. It is clear that we must find $P(A|B)$ , so we shall look to calculate $P(B|A)$ , $P(A)$ , and $P(B)$ .

I. $P(B|A)$ is the probability that one can flip a fair coin and obtain $5$ heads; this is simply $(\frac{1}{2})^5 = \frac{1}{32}$

II. $P(A)$ , the probability that the coin is fair, is $\frac{9}{10}$ , since there are $9$ fair coins and $1$ unfair coin.

III. $P(B)$ , the probability of flipping a coin (fair or unfair) and obtaining $5$ heads, is $(\frac{9}{10})(\frac{1}{32}) + (\frac{1}{10})(1) = \frac{41}{320}$ .

Thus, $P(A|B) = \frac{(\frac{1}{32})(\frac{9}{10})}{(\frac{41}{320})} = \frac{9}{41}$ , and the answer is $9+41=50$

When selecting and flipping a coin 5 times, there are 4 possible results:

1. You selected a real coin, and you got heads 5 times. The real coin gets heads half the time, and 9 of the 10 coins are real, so: $P( \mbox{real coin} \cap \mbox{5 heads} ) = \frac{9}{10} \times \frac{1}{2^5} = \frac{9}{320}$

2. You selected a real coin, but you did not get heads 5 times. The real coin gets heads half the time, so the probability of not getting all heads is 1 - P(all heads), and once again, 9 of the 10 coins are real: $P( \mbox{real coin} \cap \mbox{at least 1 tail} ) = \frac{9}{10} \times (1 - \frac{1}{2^5}) = \frac{279}{320}$

3. You selected the fake coin, and got heads 5 times. The fake coin always gets heads, and only 1 of the 10 coins is fake, so: $P( ( \mbox{fake coin} \cap \mbox{5 heads} ) = \frac{1}{10} \times 1 = \frac{1}{10}$

4. You selected the fake coin, but did not get heads 5 times. One of the 10 coins is fake, but it always get heads, so can never not get heads: $P( \mbox{fake coin} \cap \mbox{at least 1 tails} ) = \frac{1}{10} \times 0 = 0$

Now, the probability of a coin which flipped 5 heads being real is: $\frac{P ( \mbox{real coin} \cap \mbox{5 heads} ) }{P ( \mbox{5 heads} ) }\\ = \frac{P ( \mbox{real coin} \cap \mbox{5 heads} ) }{P ( \mbox{real coin} \cap \mbox{5 heads} ) + P ( \mbox{fake coin} \cap \mbox{5 heads} ) }\\ = \frac{ \frac{9}{320} }{ \frac{9}{320} + \frac{1}{10} } = \frac{9}{41}$

Greatest common divisor of 9 and 41 is 1, so they are co-prime. $\therefore a = 9, b = 41 \\ \therefore a + b = 50$

The problem asks for the probability that the drawn coin is fair given that it results in 5 heads after flipped 5 times. By definition of conditional probability, the required probability can be obtained by dividing the probability that a fair coin is drawn and getting 5 heads, by the probability of getting 5 heads.

probability that a fair coin is drawn and getting 5 heads = (9/10)(1/2)^5 = 9/320

probability of getting 5 heads = (9/10)(1/2)^5 + (1/10)(1)^5 = 41/320

Therefore, probability that the drawn coin is fair = (9/320)/(41/320) = 9/41

Thus, the answer is 9 + 41 = 50.

as mentioned there are 10 coins laid out and all of them are fair except one which flips to head every time

now,as we flip it 5 times it results in 5 flips each and results in heads so,the probability to get 5 heads (and given 10 coins is)

=(1/10)*(9/32+1)=41/320.................................(1)

now,probability that this coin is fair and then get 5 heads ------

=(1/10)(9*1/32)=9/320..........................(2)

from 1 and 2 therefore...........(9/320)/(41/320)=9/41=a/b

so,9+41=50...................the ans

The problem asks us to find the probability a randomly chosen coin is fair, given that it has tossed five heads. Let $A$ be the event that the coin we pick tosses five heads. Let $B$ be the event the coin we pick is fair. We thus want to find $P(B|A) = \frac{P(A \cap B)}{P(A)}$ (where $\cap$ denotes "and", $|$ denotes "given", and $P(A)$ denotes the probability of $A$). The probability the coin tosses five heads is $\left(\frac{1}{2}\right)^5$ for nine of the coins (the fair ones) and $1$ for the unfair coin. Thus $P(A) = \frac{9}{10} \cdot \left(\frac{1}{2}\right)^5 + \frac{1}{10} \cdot 1$ and $P(A \cap B) = \frac{9}{10} \cdot \left(\frac{1}{2}\right)^5$. This yields the conditional probability of $\frac{9}{41}$, giving us an answer of 50.

We consider the flip results of all the coins. For the 9 fair coins, they have $2^5=32$ different possibilities, of which 1 of them contains the result of 5 heads. For the remaining coin, it will always produce the result of 5 heads. Therefore, the probability that the coin is fair will be $\frac{\frac{1}{32} \times 9}{\frac{1}{32} \times 9+1} = \frac {9}{41}$ .

We apply Bayes rule to this problem. Let $Y$ be the event that the coin is a fair coin and $X$ be the event of flipping $5$ heads in $5$ flips. So, we look for $P(Y | X) = \frac{P (Y \cap X)}{P(X)}$ .

We have $P(X) = \frac{1}{10}\times 1^5 + \frac{9}{10}\times\left(\frac{1}{2}\right)^5 = \frac{1}{10} + \frac{9}{320} = \frac{41}{320}$ and $P (Y \cap X) = \frac{9}{10} \times \left(\frac{1}{2}\right)^5 = \frac{9}{320}$ .

Thus $P(Y | X) = \frac{\frac{9}{320}}{\frac{41}{320}} = \frac{9}{41}$ . Hence $a + b = 50$ .

5 heads can come in only 2 ways

1. unfair coin is selected and head comes 5 times with p1=(9/10)*(1/2^5)
2. fair coin is selected and comes 5 times(which is sure) with p2=(1/10)*(1)

so required probability=p1/(p1 + p2)=9/41.