Two balls are drawn at random from a bag containing 2 white, 3 red, 5

1.	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.
Answer» 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') = \(1-\frac{20}{91}=\frac{71}{91}\)

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.

Answer»

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') = \(1-\frac{20}{91}=\frac{71}{91}\)

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