An urn contains 3 white and 5 blue balls and a second urn contains 4 w

1.	An urn contains 3 white and 5 blue balls and a second urn contains 4 white and 4 blue balls. If one ball is drawn from each urn, what is the probability that they will be of the same colour ?
Answer» Let E : Event of drawing both the balls of same colour from the two urns E₁ : Getting 1 white ball from the first urn and 1 white ball from the second urn E₂ : Getting 1 blue ball from the first urn and 1 blue ball from the second urn Then, P(E) = P(E) + P(E₂) (∵ The two events E₁and E₂ are mutually exclusive) = {P(1 white from 1st urn) x P(1 white from 2nd urn)} + {P(1 blue from 1st urn) x P(1 blue from 2nd urn)} = \(\frac{3}{8}\times\frac{4}{8}+\frac{5}{8}\times\frac{4}{8}\) (∵ Event of drawing a ball from one urn is independent of drawing a ball from other urn) = \(\frac{12}{64}+\frac{20}{64}=\frac{32}{64}=\frac{1}{2}.\)

An urn contains 3 white and 5 blue balls and a second urn contains 4 white and 4 blue balls. If one ball is drawn from each urn, what is the probability that they will be of the same colour ?

Answer»

Let E : Event of drawing both the balls of same colour from the two urns

E₁ : Getting 1 white ball from the first urn and 1 white ball from the second urn

E₂ : Getting 1 blue ball from the first urn and 1 blue ball from the second urn

Then, P(E) = P(E) + P(E₂)

(∵ The two events E₁and E₂ are mutually exclusive)

= {P(1 white from 1st urn) x P(1 white from 2nd urn)} + {P(1 blue from 1st urn) x P(1 blue from 2nd urn)}

= \(\frac{3}{8}\times\frac{4}{8}+\frac{5}{8}\times\frac{4}{8}\)

(∵ Event of drawing a ball from one urn is independent of drawing a ball from other urn)

= \(\frac{12}{64}+\frac{20}{64}=\frac{32}{64}=\frac{1}{2}.\)

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