This question was also attempted fairly well by the candidates. Two alternative methods were open for them. Either the simple definition method of
Prob. = no of favourable
Total no of possible, which would give P {same colour} = 6C3+ 4C3
10C3
= 24 = 1 = 0.2 Or
120 5
P{same colour} = P(all read) + P(all blue) =
{ 6/10 x 5/9 x 4/8} + {4/10 x 3/9 x 2/9} = 1/5 = 0.2
The second method was the popular one. Many candidates made the mistake of interpreting P{same colour} as P{all black} x P{all blue}.
The Probability of selecting 1 black and 2 red balls = 4C1 6C2 = 0.5
10C3
