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A bag contains \[5\]white and \[8\]red balls. Two successive drawings of \[3\] balls are made such that the balls are not replaced before the second draw. Find the probability that the first drawing will give \[3\]white and the second \[3\]red balls.

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: In the given problem we have to use the concept of probability. We have to find the probability that the first drawing will give 3 white and the second 3 red balls. First, we have to find the probability that the first three balls are white in the first draw, with no replacement and then we have to find the second draw, the probability that all are red. Then we have to multiply these two results.

Complete step by step answer:
A bag contains \[5\] white and \[8\] red balls.
Total number of balls =\[13\].
Two successive drawings of 3 balls are made such that the balls are not replaced before the second draw.
We have to find the probability that the first drawing will give \[3\]white and the second \[3\]red balls.
The probability of first three balls is white in first draw, with no replacement is
\[\Rightarrow \dfrac{5}{13}\times \dfrac{4}{12}\times \dfrac{3}{11}=\dfrac{60}{1716}\]
Now the second draw, the probability that all are red is
\[\Rightarrow \dfrac{8}{10}\times \dfrac{7}{9}\times \dfrac{6}{8}=\dfrac{336}{720}\]
Thus, the probability that the first drawing will give \[3\] white and the second \[3\] red balls is
\[\Rightarrow \dfrac{60}{1716}\times \dfrac{336}{720}=\dfrac{7}{429}\]
Therefore, the probability that the first drawing will give \[3\] white and the second \[3\]red balls is \[\dfrac{7}{429}\]

Note: Students should be well aware of the basic formulae of probability. Students should be very careful while doing the calculation part for the given question. Students should have to understand the question properly. Students should have to read questions thoroughly. We should not miss any probability example in this question in second draw the probability that all are red is \[\Rightarrow \dfrac{8}{10}\times \dfrac{7}{9}\times \dfrac{6}{8}=\dfrac{336}{720}\] and not \[\Rightarrow \dfrac{8}{10}\times \dfrac{7}{9}\] .
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