Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colours.
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Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colours.
Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colours.
given: 2 white, 3 red, 5 green, 4 black
Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\)
two balls are drawn one by one, we have to find the probability that they are of different colours
total possible outcomes are 14C2
therefore n(S)= 14C2 = 91
let E be the event that all balls are of same colour
E= {WW, RR, GG, BB}
n(E)= 2C2 + 3C2 + 5C2 + 4C2 = 20
probability of occurrence is
P(E) = \(\frac{n(E)}{n(S)}\)
P(E) = \(\frac{20}{91}\)
Therefore, the probability of non-occurrence of the event (all balls are different) is
P(E’) = 1- P(E)
P(E’) = \(1-\frac{20}{91}=\frac{71}{91}\)