A bag contains N fair coins, and one coin with 'Head' on both sides.

Question

A bag contains $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\dfrac{9}{16}$, then $N$ is equal to:

Accepted Answer

Let $E_1$: selecting a fair coin, $E_2$: selecting the two-headed coin, $A$: getting a head.

$$P(E_1)=\frac{N}{N+1},\quad P(E_2)=\frac{1}{N+1},\quad P(A\mid E_1)=\frac12,\quad P(A\mid E_2)=1$$

By total probability:

$$P(A) = \frac{N}{N+1}\cdot\frac12+\frac{1}{N+1}\cdot1 = \frac{N+2}{2N+2}=\frac{9}{16}$$

$$9(2N+2)=16(N+2) \implies 18N+18=16N+32 \implies 2N=14 \implies N=7$$

Solutions · 1