Question

A die is rolled 100 times and the outcomes are recorded as below.

$\begin{array}{ccccc}\text{Outcome}& 1& \text{Even number}<\text{6}& \text{Odd number}>\text{1}& 6\\ \text{Frequency}& 15& 33& 36& 16\end{array}$

Then, which of the given options is correct?

• A. P(getting an odd number) = 0.49
• B. P(getting an even number) = 0.49
• C. P(getting an odd number) = P(getting an even number)
• D. P(getting an odd number) > P(getting an even number)

Answer 1

Answer: (B), (D)

The correct options are
B P(getting an even number) = 0.49
D P(getting an odd number) > P(getting an even number)

We know, for an event E,
$P\left(E\right)=\frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$

Number of times an odd number was rolled = 15 + 36 = 51
$\therefore \text{Probability of getting an odd number}=\frac{51}{100}=0.51$

Number of times an even number was rolled = 33 + 16 = 49
$\therefore \text{Probability of getting an even number}=\frac{49}{100}=0.49$

​​​​​​​Hence the answer.