# conditional probability or Bayes' theorem?

My try :

we can deal three different cases of removing two biased coin and 1 unbiased coin.

Individual probability is known $$\frac9{10} , \frac2{25},\frac1{50}$$

After that total is 49 as one is removed. now what to do conditional probability or Bayes' theorem ? i dont know how to implement ?

There are four possible combinations:

• Double-headed and shows heads with probability $$\frac{4}{50}\times 1= 0.08$$
• Double-tailed and shows tails with probability $$\frac{1}{50}\times 1 =0.02$$
• Fair coin and shows heads with probability $$\frac{45}{50}\times\frac12 = 0.45$$
• Fair coin and shows tails with probability $$\frac{45}{50}\times\frac12 = 0.45$$

so using $${P(A,B)}=P(A \mid B) \,P(B)$$, i.e. $$P(A \mid B) =\dfrac{P(A,B)}{P(B)}$$, and $$P(B)= P(A,B)+P(A^c,B)$$, you get

(a) Given it shows heads, the conditional probability of being double-headed is $$\frac{0.08}{0.08+0.45}$$

(b) Given it shows heads, the conditional probability of being fair is $$\frac{0.45}{0.08+0.45}$$

(c) Given it shows tails, the conditional probability of being fair is $$\frac{0.45}{0.02+0.45}$$

Here's how I think about these problems. There are $$50$$ coins with $$2$$ sides each, or $$100$$ possible outcomes, each equally likely. Of those possible outcomes, $$53$$ result in heads and $$47$$ result in tails.

Out of the $$53$$ possibilities that result in heads, $$8$$ of them come from a two-headed coin and the other $$45$$ come from fair coins. So if your result is heads, the probability of a two-headed coin is $$\frac{8}{53}$$ and the probability of a fair coin is $$\frac{45}{53}.$$ If the result is tails, the probability of a fair coin is $$\frac{45}{47}$$.

• +1 Indeed it is very handsome to look at sides instead of coins in problems like this. – drhab Jan 20 at 9:28