Let the random variable X represent the positive difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. Then probability $P(X ≤ 3)$ is equal to.

25/32

The correct answer is Option (3) → 25/32

Let X = |H − T| for 6 tosses.

Since H + T = 6,

$X = |2H - 6|$

Requirement:

$X \le 3 \Rightarrow |2H - 6| \le 3$

$-3 \le 2H - 6 \le 3$

$3 \le 2H \le 9$

$1.5 \le H \le 4.5$

Thus possible integer values of H: 2, 3, 4

Total outcomes = 64

Favourable outcomes:

H = 2 ⇒ number of ways = 15

H = 3 ⇒ number of ways = 20

H = 4 ⇒ number of ways = 15

Total favourable = 15 + 20 + 15 = 50

Probability:

$P(X \le 3) = \frac{50}{64} = \frac{25}{32}$