Question

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

Answer 1

Let E1=5 or 6 appears on a die,E2=1,2,3 or 4 appears on a die and A=A head appears on the coin.
Now P(E1)=$\frac{2}{6}$=$\frac{1}{3}$,P(E2)=$\frac{4}{6}$=$\frac{2}{3}$
Now P(A|E1)Probability of getting a head on tossing a coin three times,
When E1 has already occurs=P(HTT)or P(THT)or P(TTH)
$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{3}{8}$
P(A|E2)=Probability of getting a head on tossing a coin once,
When E2 has already occured=$\frac{1}{2}$
∴P(there is exactly one head given that 1,2,3 or 4 appears on a die)
$P(E_2|A)=\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$
=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}$$=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}$$=\frac{24}{3}\times11=$\frac{24}{33}$=$\frac{8}{11}$