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There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls. - Mathematics

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Sum

There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.

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Solution

There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in 5C3= `(5 xx 4 xx 3)/(1 xx 2 xx 3)` = 10 ways.

∴ n(S) = 10
Let A be the event that 2 balls are red and 1 ball is black
2 red balls can be drawn out of 2 red balls in 2C2 = 1 way and 1 black ball can be drawn out of 3 black balls in 3C1 = 3 ways.
∴ n(A) = 2C2 × 3C1 = 1 × 3 = 3

∴ P(A) = `("n"("A"))/("n"("S")) = 3/10`
Let B be the event that 1 ball is red and 2 balls are black 
1 red ball out of 2 red balls can be drawn in 2C1 = 2 ways and 2 black balls out of 3 black balls can be drawn in 3C2 = `(3 xx 2)/(1 xx 2)` = 3 ways.

∴ n(B) = 2C1 × 3C2 = 2 × 3 = 6

∴ P(B) = `("n"("B"))/("n"("S")) = 6/10`

Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B) = P(A) + P(B)

= `3/10 + 6/10`

= `9/10`

Concept: Introduction of Probability
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 7 Probability
Miscellaneous Exercise 7 | Q 4 | Page 110

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