Question

A coin is tossed and a die is rolled one after the other. The probability of flipping a tail and seeing a 4 or flipping a head and seeing a 6 is .

• A. $\frac{1}{6}$
• B. $\frac{2}{3}$
• C. $\frac{1}{12}$

Answer 1

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

We represent $\text{H}$ and $\text{T}$ for heads and tails, respectively.

Moreover, we represent the compound events of flipping a tail and rolling a $4$ by $\text{T4}$ and flipping a head and rolling a $6$ by $\text{H6}$.

$\therefore \text{P(T4 or H6)=P(T4)+P(H6)}$

Now,
$\text{P(T4)}$
$=\text{P(tail and 4)}$
$=\text{P(tail)}\times \text{P(4)}$
$=\text{P(T)}\times \text{P(4)}$
$=\frac{1}{2}\times \frac{1}{6}$
$=\frac{1}{12}$

Next,
$\text{P(H6)}$
$=\text{P(head and 6)}$
$=\text{P(head)}\times \text{P(6)}$
$=\text{P(H)}\times \text{P(6)}$
$=\frac{1}{2}\times \frac{1}{6}$
$=\frac{1}{12}$

$\therefore \text{P(T4 or H6)=P(T4)+P(H6)}$

$\Rightarrow \text{P(T4 or H6)}=\frac{1}{12}+\frac{1}{12}$

​​​​​​​$\Rightarrow \text{P(T4 or H6)}=\frac{2}{12}=\frac{1}{6}$

Hence, the probability of flipping a tail and seeing a 4 or flipping a head and seeing a 6 is $\frac{1}{6}$.