Question

Lewis rolls a number cube that has sides labelled 1 to 6 and then flips a coin. The probability that Lewis rolls an even number and flips a head is .

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{4}$

Sample Space Diagram:

	1	2	3	4	5	6
Hesads (H)	H1	H2	H3	H4	H5	H6
Tails (T)	T1	T2	T3	T4	T5	T6

Total number of possible outcomes= $12=2\times 6=\text{row}\times \text{column}$

Let $\text{A}$ is an event of rolling an even number (E) and flipping a head (H).

Favorable Outcomes for event $\text{A}$= $\text{{H2, H4, H6}}$

$\therefore$ Number of favorable outcomes to the event $\text{A}$ = $3$

$\text{P(A)}=\frac{\text{Number of outcomes favorable to A}}{\text{Total number of possible outcomes}}$

$\Rightarrow \text{P(A)}=\frac{3}{12}=\frac{1}{4}$


$\underset{–}{\text{Alternative method}}$

Suppose, $\text{P(E)}$, $\text{P(H)}$ and $\text{P(E and H)}$ are the probabilities of rolling an even number on die, flipping a head on coin and rolling an even number and flipping a head, respectively.

We know,
$\text{P(E and H) = P(E)}\times \text{P(H)}$

Now, to find $\text{P(E and H)}$, we need to find out $\text{P(E)}$ and $\text{P(H)}$.

Total number of possible outcomes for rolling a die= $6$

Favorable outcomes to the event $\text{E}$= $\text{{2, 4, 6}}$
$\therefore$ Number of outcomes favorable to event $\text{E}$= $3$

$\therefore \text{P(E)}=\frac{\text{Number of outcomes favorable to event E}}{\text{Total number of possible outcomes}}$
$\Rightarrow \text{P(E)}=\frac{3}{6}=\frac{1}{2}$

Flipping a coin has two possibilities= $\text{{H, T}}$
$\therefore \text{Probability of flipping head= P(H)}=\frac{1}{2}$

Finally,
Probability that Lewis rolls an even number and flips a head = $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$


$\therefore$ The probability of rolling an even number and flipping a head is $\frac{1}{4}$.