Consider the following cases :


①First toss : 1 head 1 tail , game ends

② First toss : 2 heads, game continues.
Second toss : 2 heads and 1 tail

③First toss : 2 heads, game continues.
Second toss : 1 head and 2 tails

④First toss : 2 heads, game continues.
Second toss : 3 heads

⑤First toss : 2 heads, game continues.
Second toss : 3 tails

⑥First toss : 2 tails, game continues.
Second toss : 2 heads and 1 tail

⑦First toss : 2 tails, game continues.
Second toss : 1 head and 2 tails

⑧First toss : 2 tails, game continues.
Second toss : 3 heads

⑨First toss : 2 tails, game continues.
Second toss : 3 tails


P(X = 2) covers the probabilities for case ⑤ and case ⑥



For case ⑤, probability

= (½)² × (½)³
= 1/32

(For the first toss : ½ for first head × ½ for second head.

For second toss : ½ for each of the three heads)

For case ⑥,

Probability

= (½)² × (½)³ × 3

= 3/32


(For first toss : ½ for first tail and ½ for second tail.

For second toss : ½ for one tail , ½ for each of the two heads.

But 3 cases are possible (HHT, HTH, THH) so multiply by 3)


Total probability = 1/32 + 3/32
= 4/32
= ⅛

(Shown)