Question

Fredrick rolls a number cube that has sides labelled $1$ to $6$ and then flips a coin. What is the the probability that Fredrick rolls an odd number and flips a head?

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

Answer 1

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

Firstly, let's draw the sample space diagram for the given experiment.

Sample Space Diagram:

Total number of possible outcomes $=2\times 6=12$

$\underset{–}{\text{Need to Find}:}$
Probability of rolling an odd number and filliping a head

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

Favorable Outcomes for event $\text{A}$ $=\text{{H1, H3, H5}}$ as depicted in below diagram,
$\therefore$ 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, $\text{P(O)}$, $\text{P(H)}$ and $\text{P(E and H)}$ are the probabilities of rolling an odd number on die, flipping a head on coin and rolling an odd number and flipping a head, respectively.

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

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

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

Favorable outcomes to the event $\text{O}$ $=\text{{1, 3, 5}}$
$\therefore$ Number of outcomes favorable to event $\text{O}$ $=3$

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

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

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

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

Hence, option (d.) is the correct choice.