Question

A coin is tossed and a die is rolled one after the other. What is the probability of flipping a tail and seeing a 4?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{8}$
• D. $\frac{1}{12}$

Answer 1

The correct option is D $\frac{1}{12}$

$\bullet \underset{–}{\text{Solve using sample space diagram}}$


In the very first column, we have two possibilities (heads or tails) for flipping a coin.

In the very first row, we have all the possibilities $(1,2,3,4,5,6)$ for rolling a six-sided fair die.

Here, the compound event is flipping a tail and rolling a $4$, which is $\text{T4}$ in the sample space diagram.

Total number of possible outcomes for flipping a coin and rolling a die one after the other= $12$
Number of favourable outcomes for the compound event of flipping a tail and rolling a $4$= 1

$\therefore \text{P(T4)}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}=\frac{1}{12}$

$\bullet \underset{–}{\text{Solve using tree diagram}}$


Total number of possibilities= $12$
For $\text{T4}$, number of favorable outcomes= $1$

$\therefore \text{P(T4)}=\frac{1}{12}$

$\bullet \underset{–}{\text{Solve using multiplication of single events}}$

Here, flipping a tail and rolling a $4$ are independent compound events.

$\therefore \text{P(T4)= P(T)}\times \text{P(4)}$

Number of favorable outcomes for flipping a tail= $1$
Total number of possible outcomes for flipping a coin= $2$
Hence, $\text{P(T)}=\frac{1}{2}$

Number of favorable outcomes for rolling a $4$= $1$
Total number of possible outcomes for rolling a die= $6$
Hence, $\text{P(4)}=\frac{1}{6}$

$\therefore \text{P(T4)= P(T)}\times \text{P(4)}$

$\Rightarrow \text{P(T4)}=\frac{1}{2}\times \frac{1}{6}$

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

$\therefore$ Using all these above three methods, we get the probability of flipping a tail and rolling a $4$ is $\frac{1}{12}$.