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 $P(X \leq 3)$ is equal to

When a coin is tossed 6 times, let $h$ be the number of heads and $t$ be the number of tails.

$h + t = 6$

The random variable $X = |h - t|$ represents the positive difference.

---

Since $t = 6 - h$:

$X = |h - (6 - h)|$

$X = |2h - 6|$

$X = 2|h - 3|$

---

For each possible number of heads:

$h = 0$: $X = 2|0-3| = 6$

$h = 1$: $X = 2|1-3| = 4$

$h = 2$: $X = 2|2-3| = 2$

$h = 3$: $X = 2|3-3| = 0$

$h = 4$: $X = 2|4-3| = 2$

$h = 5$: $X = 2|5-3| = 4$

$h = 6$: $X = 2|6-3| = 6$

The possible values of $X$ are: $0, 2, 4, 6$

---

For $X \leq 3$, the values that satisfy this condition are $X = 0$ and $X = 2$.

---

For a fair coin tossed 6 times, the total number of outcomes is $2^6 = 64$.

The probability of getting exactly $h$ heads is $\dfrac{C(6,h)}{64}$.

$X = 0$ occurs when $h = 3$:

$C(6,3) = 20$

$P(X = 0) = \dfrac{20}{64}$

---

$X = 2$ occurs when $h = 2$ or $h = 4$:

$C(6,2) = 15$

$C(6,4) = 15$

$P(X = 2) = \dfrac{15 + 15}{64} = \dfrac{30}{64}$

---

$P(X \leq 3) = P(X = 0) + P(X = 2)$

$P(X \leq 3) = \dfrac{20}{64} + \dfrac{30}{64}$

$P(X \leq 3) = \dfrac{50}{64}$

$P(X \leq 3) = \dfrac{25}{32}$