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A box contains 4 coins – 3 fair coins and 1 biased coin for which P(heads) = 0.8. A coin is picked at random and tossed 6 times. It shows 5 heads. Find the probability this coin is fair.

Would you tell me how to solve this question step by step?

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  • $\begingroup$ To start: what are the odds that you'll get 5 heads out of 6 tosses with a fair coin? $\endgroup$ – Théophile Aug 12 at 14:48
  • $\begingroup$ Baye's Formula is good way to go about solving this problem. Are you familiar with what it states? $\endgroup$ – questionasker Aug 12 at 18:58
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I would proceed like this.

  1. Write clearly the conditional probability being asked for in the question. We want the following: the probability the coin is fair, given that it comes up heads 5 times out of 6. Let $A$ be the event that the coin is fair and let $B$ the event that the coin comes up heads 5 of 6 times.
  2. Conditional probability is evaluated as $P(A|B) = \frac{P(AB)}{P(B)}$. Bayes formula gives us $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$. The probability of event $B$ occurring depends on the coin being flipped. So we express $P(B)$ using the "law of total probability" as follows: $P(B) = P(B|A)P(A) + P(B|\bar{A})P(\bar{A})$. So our expression becomes $P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\bar{A})P(\bar{A})}$.
  3. Now, $A$ is the event that the coin is fair. So $P(A) = \frac{3}{4}$ and $P(\bar{A}) = \frac{1}{4}$
  4. The number of times that the coin comes up heads in 6 flips, is a random variable with binomial distribution with $n = 6$. $p = \frac{1}{2}$ when the coin is fair and $p = \frac{4}{5}$ otherwise. Use this to calculate $P(B|A)$ and $P(B|\bar{A})$.

In summary, the concepts used here that you'll want to familiarize yourself with are "the law of total probability", "bayes formula/theorem", "binomial distribution". I promise this will all feel intuitive once you're familiar.

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Firstly, we define the events:

  • $F$: the dice is fair
  • $B$: the dice is biased
  • $5H$: five heads out of six tosses

With those events defined, the event we are looking for is $F | 5H$. We can use the Bayes' rule

$$P(F|5H) = \frac{P(5H|F)\cdot P(F)}{P(5H)}$$

Now we have to find every probability in the RHS:

  • $P(F) = \frac{3}{4}$ (since the coined is originally picked at random)
  • $P(5H|F) = {6\choose 5} (\frac{1}{2})^5 (1 - \frac{1}{2})^1$ (binomial distribution)
  • $P(5H) = P(5H|F)\cdot P(F) + P(5H|B)\cdot P(B)$ (Law of total probability)

Simplifying the last two probabilities gives $P(5H|F) = \frac{6}{64}$ and

$P(5H) = \frac{6}{64} \cdot \frac{3}{4} + (5\cdot\frac{8}{10}^5 \cdot\frac{2}{10}) \cdot \frac{1}{4} $

Plug the numbers in the Bayes' rule and you got the answer.

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  • $\begingroup$ Thank you everyone for your detailed and clear answers (and comments). One thing I cannot get is the value of P(5H|F). I wonder why it is not 3/32 because I think the probability of getting 5 heads in 6 flips can be found by 6C5 divided by 64 (2 to the power of 6), which is 3/32. $\endgroup$ – Peppers Aug 14 at 8:44
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    $\begingroup$ You are correct and I am an idiot. $6 \choose 5$ is 6 and not 5 as I wrote. I will edit it. $\endgroup$ – cgss Aug 14 at 8:50

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