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)
\(P(E_2|A)=\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}\)\(=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}\)\(=\frac{24}{3}\times11\)=\(\frac{24}{33}\)=\(\frac{8}{11}\)