Question

**Assertion (A):** The probability that a number selected at random from the numbers $1, 2, 3, ......, 15$ is a multiple of 4, is $\frac{1}{3}$. **Reason (R):** Two different coins are tossed simultaneously. The probability of getting at least one head is $\frac{3}{4}$.

Accepted Answer

D. - Multiples of 4 between 1 and 15 are $\{4, 8, 12\}$ (3 numbers out of 15). Probability $= \frac{3}{15} = \frac{1}{5} 
eq \frac{1}{3}$. So Assertion (A) is false. - When two coins are tossed, sample space is $\{HH, HT, TH, TT\}$. At least one head occurs in $\{HH, HT, TH\}$ (3 outcomes). Probability $= \frac{3}{4}$. So Reason (R) is true.

Assertion-Reason: Probability of multiples and coin toss

Explanation

Practise this under board exam conditions