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Question
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
- both balls are red.
- first ball is black and second is red.
- one of them is black and other is red.
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Solution
Let R = the event of drawing a red ball; B = the event of drawing a black ball
i. Probability of getting a red ball in the first draw P(R) = `8/(10 + 8) = 8/18 = 4/9`
Because the ball is put back again.
∴ The probability of getting a red ball in the second draw P(R) = `4/9`
∴ The probability of both balls being red = P(R). P(R) = `4/9 xx 4/9 = 16/81`
ii. Probability of getting a black ball in the first draw P(B) = `10/18 = 5/9`
The probability of getting a red ball in the second draw P(R) = `4/9`
∴ P(first black and second red) = P(B) . P(R) = `5/9 xx 4/9 = 20/81`
iii. P(one black and one red) = P(first black and second red) + P(first red and second black)
= `5/9 . 4/9 + 4/9 . 5/9`
= `40/81`
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