Question

Suppose, a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die?

Answer 1

Let $E_1$ the event that 5 or 6 is shown on die and $E_2$ the event that 1,2,3, or 4 is shown on die.

Then $E_1$ and $E_2$ are mutually exclusive and exhaustive events. and n$E_1$=2, n$E_2$ =4

Also, n(S)= 6

$\therefore P\frac{E}{E_1}$ = P (exactly one head show up when coin is tossed thrice)

=P(HTT,THT,TTH)= $\frac{3}{8}$ $\because$ total number of events =$2^3=8)$

$P\frac{E}{E_2}$ = P (head shows up when coin is tossed once) = $\frac{1}{2}$

The probability that the girl threw, 1,2,3 or 4 with the die, if she obtained exactly one head, is given by $P\frac{E_2}{E}$

By using Baye's theorem, we obtain

$P\left(\frac{E_2}{E}\right)=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}=\frac{\frac{1}{2}\times \frac{2}{3}}{\frac{1}{2}\times \frac{2}{3}+\frac{3}{8}\times \frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}=\frac{8}{8+3}=\frac{8}{11}$.