Explanation

43/96

Let the girls be A, B and C

P(A pass) = $2/3$ , P'(A)

P(A fails) = $1 -$ P(A pass) = $1 - 2/3 = 1/3$

P(B pass) = $5/8$

P(B fails) = $3/8 =$ P(B)

P(C pass) = $3/4$

P(C fails ) = $1/4 =$ P(C)

.: P(only two of the girls pass)

P(A) × P(B) × P'(C) OR

P(A) P(C) × P'(B) or

P(B) P(C) × P'(A)

= $(2/3 * 5/8 * 1/4)$ + $(2/3 × 3/4 × 3/8)$ + $(3/8 × 3/4 × 1/3 )$

= $10/96 + 18/96 + 15/96$

$(10 + 18 + 15) / 96 = 43/96$

43/96

Let the girls be A, B and C

P(A pass) = $2/3$ , P'(A)

P(A fails) = $1 -$ P(A pass) = $1 - 2/3 = 1/3$

P(B pass) = $5/8$

P(B fails) = $3/8 =$ P(B)

P(C pass) = $3/4$

P(C fails ) = $1/4 =$ P(C)

.: P(only two of the girls pass)

P(A) × P(B) × P'(C) OR

P(A) P(C) × P'(B) or

P(B) P(C) × P'(A)

= $(2/3 * 5/8 * 1/4)$ + $(2/3 × 3/4 × 3/8)$ + $(3/8 × 3/4 × 1/3 )$

= $10/96 + 18/96 + 15/96$

$(10 + 18 + 15) / 96 = 43/96$