Question

Online game survey is done for $180$ people. If we select a person randomly, then:


Statement I: $P(<=18$ years and plays online games) $=14$
Statement II: $P(18–25$ years and do not play online games) = $19$
Select the correct option:

• A. Both the statements are correct.
• B. Statement I is correct and statement II is incorrect.
• C. Both the statements are incorrect.
• D. Statement I is incorrect and statement II is correct.

Answer 1

The correct option is D Statement I is incorrect and statement II is correct.

Let us analyze and calculate the probability for both the given statements.
Statement I:
Probability: Probability is defined as the likelihood of something occurring or the chance of something happening.

So, $\text{Probability(event)}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Let event A be, a person selected is $\le 18$ years old and plays online games.
Using the table, we have
Total number of people $=80+35+45+20=180$
Number of people who are $\le 18$ years old and plays online games $=80$

So, $\text{P(A)}=\frac{\text{Number of people who are ≤ 18 years old and plays online games}}{\text{Total number of people}}=\frac{80}{180}$ or $\frac{4}{9}\ne \frac{1}{4}$

$\Rightarrow$ Statement I is incorrect.

Statement II:
Probability: Probability is defined as the likelihood of something occurring or the chance of something happening.

So, $\text{Probability(event)}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Let event B be, a person selected is $18–25$ years old and does not play online games.
Using the table, we have
Total number of people $=80+35+45+20=180$
Number of people who are $18–25$ years old and does not play online games $=20$

So, $\text{P(B)}=\frac{\text{Number of people who are 18-25 years old and does not play online game}}{\text{Total number of people}}=\frac{20}{180}$ or $\frac{1}{9}$

$\Rightarrow$ Statement II is correct.