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A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, one is white and other is black?

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Question

A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, one is white and other is black?

Sum
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Solution

Total number of balls = 10 + 15 = 25
Let S be event that two balls are drawn at random without replacement in succession
∴ n(S) = `""^25"C"_1xx""^24"C"_1` = 25 × 24

Let B be the event that one ball is white and other is black.
In this case, either 1st ball drawn is white and 2nd is black or 1st is black and 2nd is white.
First white ball can be drawn from 10 white balls in 10C1 ways and second black ball can be drawn from 15 black balls in 15C1 ways.
Similarly, first black ball from 15 black balls can be drawn in 15C1 ways and second white ball from 10 white balls can be drawn in 10C1 ways.

∴ n(B) = `""^10"C"_1""^15"C"_1+""^15"C"_1 ""^10"C"_1`

∴ P(B) = `("n"("B"))/("n"("S"))=(10xx15)/(25xx24)+(15xx10)/(25xx24)`

= `150/(25xx24)+150/(25xx24)`

= `300/(25xx24)`

= `1/2`

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Chapter 7: Probability - Exercise 7.4 [Page 108]

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