Question

Suppose a girl throws a die. If she gets a $5$ or $6$, she tosses a coin $3$ 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?

• A. $\frac{1}{9}$
• B. $\frac{5}{16}$
• C. $\frac{2}{13}$
• D. $\frac{8}{11}$

Answer 1

Answer: (D)

$\frac{8}{11}$

Let $E_1$ be the event that the outcome on the die is $5$ or $6.$

Let $E_2$ be the event that the outcome on the die is $1,2,3$ or $4.$

Therefore,

$P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

$P\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$

Let $A$ be the event of getting exactly one head. Then,

$P\left(\frac{A}{E_1}\right)=\frac{3}{8}$

$P\left(\frac{A}{E_2}\right)=\frac{1}{2}$

Using Baye's theorem,

$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)\cdot P\left(\frac{A}{E_2}\right)}{P\left(E_1\right)\cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right)\cdot P\left(\frac{A}{E_2}\right)}$

$P\left(\frac{E_2}{A}\right)=\frac{\frac{2}{3}\cdot \frac{1}{2}}{\frac{1}{3}\cdot \frac{3}{8}+\frac{2}{3}\cdot \frac{1}{2}}$

$=\frac{1}{\frac{3}{8}+1}$ $=\frac{1}{\frac{11}{8}}$ $=\frac{8}{11}$.