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A box contains 5 red and 4 white marbles...

A box contains 5 red and 4 white marbles. Two marbles are drawn successively from the box without replacement and the second drawn marble drawn is found to be white. Probability that the first marble is also white is

Text Solution

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Let A be the event that first marble is white.
let B be the event that second marble is white.
Let C be the event that first marble is not white.
Applying Bayes theorem
`P(A/B)=(P(B/A)P(A))/(P(B/A)P(A)+P(B/C)P(C))`
therefore,`P(A/B)= (4/9 xx 3/8)/((4/9 xx 3/8)+(5/9 xx 4/8))=12/32=3/8`
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