Each of the `n`
urns contains 4 white and 6 black balls. The `(n+1)`
th urn contains 5 white and 5 black balls. One of the `n+1`
urns is chosen at random and two balls are drawn from it
without replacement. Both the balls turn out to be black. If the probability
that the `(n+1)`
th urn was chosen to draw the balls is 1/16, then
find the value of `n`
.
Text Solution
Verified by Experts
Let `E_(1)` denoter rthe evetn that one of the first n urns is chosen and `E_(2)` donate the evetn that (n + 1)th urn is selected. A denotes the event that two balls drawn are black. Then, `P(E_(1))=n//(n+1),P(E_(2))=1//(n+1),` `P(A//E_(1))=""^(6)C_(2)//^(10)C_(2)=1//3` and `P(A//E_(1))=""^(6)C_(2)//^(10)C_(2)=2//9.` Usign Bayes's theorem, we have `P(E_(2)//A)=(P(E_(2))P(A//E_(2)))/(P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2)))` `or 1/16=(((1)/(n+1))2/9)/(((n)/(n+1))((1)/(3))+((1)/(n+1))((2)/(9)))=(2)/(3n+2)or n=10`
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