Correct Answer – Option 3 : 63/703

Given:

Black balls = 13

White balls = 12

Red balls = 15

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = Number of ways of selecting 4 balls out of 40 ball

⇒ n(S) = ⁴⁰C₄ = 40!/((40 – 4)! × 4!)

⇒ (40!/36! × 4!) = (40 × 39 × 38 × 37)/(4 × 3 × 2)

Let E be the event of getting 2 black and 2 red ball

n(E) = ¹³C₂ × ¹⁵C₂

⇒ (13 × 12)/2 × (15 × 14)/2

P(E) = n(E)/n(S)

⇒ P(E) = \(\frac{{\frac{{13{\rm{\;}} × {\rm{\;}}12}}{2} × \frac{{{\rm{\;}}\left( {15{\rm{\;}} × {\rm{\;}}14} \right)}}{2}}}{{\frac{{40{\rm{\;}} × {\rm{\;}}39{\rm{\;}} × {\rm{\;}}38{\rm{\;}} × {\rm{\;}}37}}{{4{\rm{\;}} × {\rm{\;}}3{\rm{\;}} × {\rm{\;}}2}}}}\)

⇒ (13 × 12 × 15 ×14 × 3 × 2)/(40 × 39 × 38 × 37)

⇒ 63/703

 ∴ The probability of getting 2 black ball and 2 red ball is 63/703