If from each of the three boxes containing 3 red and 1 green; 2 red and 2 green; 1 red and 3 green balls, one ball is drawn at random, then the probability that 2 red and 1 green ball we drawn, is given by:

Concept:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects, is given by: \(\rm Probability =\dfrac{^{p}C_{k}}{^{n}C_{k}}\).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

  P[(A and B) or (B and C)]

  = P(A and B) + P(B and C)

  = [P(A) × P(B)] + [P(C) × P(D)]

  ('and' means '×' and 'or' means '+')

Calculation: 

Let box A, box B and box C be the three boxes respectively.The given event is a Compound Event which can be written as:

[(1G from box A) and (1R from box B) and (1R from box C)]

or [(1R from box A) and (1G from box B) and (1R from box C)]

or [(1R from box A) and (1R from box B) and (1G from box C)]

∴ \(\rm P = \left(\dfrac{^{1}C_{1}}{^{4}C_{1}}\times \dfrac{^{2}C_{1}}{^{4}C_{1}}\times \dfrac{^{1}C_{1}}{^{4}C_{1}}\right)+\left(\dfrac{^{3}C_{1}}{^{4}C_{1}}\times \dfrac{^{2}C_{1}}{^{p}C_{1}}\times \dfrac{^{1}C_{1}}{^{p}C_{1}}\right)+\left(\dfrac{^{3}C_{1}}{^{4}C_{1}}\times \dfrac{^{2}C_{1}}{^{4}C_{1}}\times \dfrac{^{3}C_{1}}{^{4}C_{1}}\right)\)

⇒ \(\rm P = \left(\dfrac{1}{4} \times \dfrac{2}{4} \times\dfrac{1}{4} \right) + \left(\dfrac{3}{4} \times \dfrac{2}{4} \times\dfrac{1}{4} \right) + \left(\dfrac{3}{4} \times \dfrac{2}{4} \times\dfrac{3}{4} \right)\)

⇒ \(\rm P =\dfrac{2}{64} + \dfrac{6}{64} +\dfrac{18}{64}\)

⇒ \(\rm P =\dfrac{26}{64}\)

⇒ \(\rm P =\dfrac{13}{32}\).