# Probability of picking a biased coin

Suppose you have a bag of 100 coins of which 1 is biased with both sides as Heads. You pick a coin from the bag and toss it three times. The result of all three tosses is Heads. What is the probability that the selected coin is biased?

P(selecting a biased coin) = 1/100
P(getting a head thrice with the biased coin) = 1

P(selecting an unbiased coin) = 99/100
P(getting a head thrice with the unbiased coin) = 1/8

P(selecting a biased coin|coin toss resulted in 3 heads) =
P(selecting a biased coin and getting heads thrice)/
[P(selecting a biased coin and getting heads thrice) +
P(selecting an unbiased coin and getting heads thrice)]

= (1/100)/[(1/100) + (99/800)]
= (1/100)/(107/800)
= 8/107
= 0.0747


Is this correct? Thanks.

• Is this a homework question? – Carlos Accioly Feb 19 '13 at 12:19
• @CarlosAccioly This question was posed to me in an examination on Sunday. Just wanted to verify my answer. – tejas_kale Feb 19 '13 at 12:44
• Hi tejas, a standard-level question on an exam - even if you're pursuing it for self study - falls under the scope of the homework tag. – Glen_b Feb 20 '13 at 2:16
• @Glen_b Got your point. I thought that Carlos' question was to ensure that the poster is not cheating with his homework. Thanks for the info. – tejas_kale Feb 20 '13 at 11:48

Your answer is right. The solution can be derived using Bayes' Theorem:

$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

You want to know the probability of $P(\text{biased coin}|\text{three heads})$.

What do we know?

There are $100$ coins. $99$ are fair, $1$ is biased with both sides as heads.

With a fair coin, the probability of three heads is $0.5^3 = 1/8$.

The probability of picking the biased coin: $P(\text{biased coin}) = 1/100$.

The probability of all three tosses is heads: $P(\text{three heads}) = \frac{1 \times 1+ 99 \times \frac{1}{8}}{100}$.

The probability of three heads given the biased coin is trivial: $P(\text{three heads}|\text{biased coin}) = 1$.

If we use Bayes' Theorem from above, we can calculate

$$P(\text{biased coin}|\text{three heads}) = \frac{1 \times 1/100}{\frac{1 + 99 \times \frac{1}{8}}{100}} = \frac{1}{1 + 99 \times \frac{1}{8}} = \frac{8}{107} \approx 0.07476636$$

• Cool! Cool! Cool! – Behacad Feb 19 '13 at 14:33

Here is a write that describes something very similar to that. The Bayes approach is the right way to proceed.

We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.

In general, P(biased coin|k heads) = (2^k)/[(2^k) + 99] Where k is no of consective heads

so if the trick coin was tossed 3 times (2^3)/[(2^3) + 99] = 8/(8+99) = 8/107 = 0.07