Question

A coin is flipped and a fair dice is rolled simultaneously. John wins the game if he gets tail on the coin and $2$ on the dice or head on the coin and $6$ on the dice. What would be the value of the probability of John winning the game?

• A. $\frac{1}{144}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$

Answer 1

The correct option is B $\frac{1}{6}$

The value of the probability of getting tail on the coin and $2$ on the dice $=P\left(tailand2\right)$
Since the given two events are compound events,
$P\left(tailand2\right)=P\left(tail\right)\times P\left(2\right)$
$P\left(tail\right)=\frac{1}{2}$
$P\left(2\right)=\frac{1}{6}$
$P\left(tailand2\right)=\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$

The value of the probability of getting head on the coin and $6$ on the dice $=P\left(headand6\right)$
Since the given two events are compound events,
$P\left(headand6\right)=P\left(head\right)\times P\left(6\right)$
$P\left(head\right)=\frac{1}{2}$
$P\left(6\right)=\frac{1}{6}$
$P\left(headand6\right)=\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$

$P\left(\text{John wins}\right)=P\left[\text{(tail and 2) or (head and 6)}\right]$
$\Rightarrow P\left(\text{John wins}\right)=P\left(\text{tail and 2}\right)+P\left(\text{head and 6}\right)$
$\Rightarrow P\left(\text{John wins}\right)=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$