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A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 white balls and second drawing

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प्रश्न

A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 white balls and second drawing will give 3 red balls.

योग
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उत्तर

Total number of balls = 8 + 5 = 13
3 balls can be drawn out of 13 balls in 13C3 ways.
∴ n(S) = 13C3
Let A be the event that all 3 balls drawn are white.
3 white balls can be drawn out of 5 white balls in 5C3 ways.
∴ n(A) = 5C3

∴ P(A) = `("n"("A"))/("n"("S")) = (""^5"C"_3)/(""^13"C"_3) = (5 xx 4 xx 3)/(13 xx 12 xx 11) = 5/143`

After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls.
Let B be the event that the second draw of 3 balls are red.
∴ Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by

`"P"("B"/"A") = (""^8"C"_3)/(""^10"C"_3) = (8 xx 7 xx 6)/(10 xx 9 xx 8) = 7/15`

∴ Required probability = P(A ∩ B)

= `"P"("A")."P"("B"/"A")`

= `5/143 xx 7/15`

= `7/429`

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अध्याय 7: Probability - Miscellaneous Exercise 7 [पृष्ठ ११०]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 11 Standard Maharashtra State Board
अध्याय 7 Probability
Miscellaneous Exercise 7 | Q 13 | पृष्ठ ११०

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