There are two bags A and B . Bag A contains n white balls and 2 black balls while bag B contains 2 white balls and n black balls . One of the two bags is selected at random and two balls are drawn from it without replacement . We know that if both the balls drawn are white balls, the probability of bag A being selected to draw the balls is 7 6 . Find the value of n .
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Probability of selecting bag A and selecting 2 white balls from bag A are two different things. First you select the bag then two whites.I think the question is incomplete.
let E1 denotes the event that bag A is selected and E2 denotes the event that bag B is selected .Let E be the event that two balls drawn are white . We have
P(E1)=1/2=P(E2) P(E|E1)=n(n-1)/(n+2)(n+1) and P(E|E2)=2/(n+2)(n+1)
Using Baye's Theorem ,The probability that the two white balls drawn are from the bag A ,is given by:
P(E1|E)=(P(E1) P(E|E1))/(P(E1) P(E|E1)+P(E2)*P(E|E2))=6/7
=n(n-1)/(n(n-1)+2)=6/7
=> n=4,-3
since n cannot be negative ,we get
=> N=4
I started to write a solution but I accidentally hit "keep private" so now I can't edit it or anything bc of dumbness.
Anyways, I'll write it here.
Let us first consider the probability of choosing 2 whites from each bag.
Bag A:
2B and nW
The probability of choosing two whites is therefore ( 2 n + 2 ) ( 2 n )
Bag B:
nB and 2W
The probability of choosing two whites is therefore ( 2 n + 2 ) ( 2 2 )
If you chose two whites, the probability that you chose bag A is 7 6 , and then clearly the probability that you chose bag B is 7 1 . The probability of choosing A is therefore 6 times that of choosing B.
Therefore,
( 2 n + 2 ) ( 2 n ) = 6 × ( 2 n + 2 ) ( 2 2 )
( 2 n ) = 6
2 n 2 − n = 6
n 2 − n − 1 2 = 0
n = 4 or n = − 3
− 3 < 0 therefore is inadmissable.
And so n = 4
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The probability of choosing 2 white balls from Bag A is:
2 1 ( n + 2 n ) ( n + 1 n − 1 ) = 2 ( n + 1 ) ( n + 2 ) n ( n − 1 )
The probability of choosing 2 white balls from Bag B is:
2 1 ( n + 2 2 ) ( n + 1 1 ) = ( n + 1 ) ( n + 2 ) 1
The probability of selecting Bag A from the 2 bags is:
B a g A + B a g B B a g A = 7 6
By substituting, we get:
2 ( n + 1 ) ( n + 2 ) n ( n − 1 ) + ( n + 1 ) ( n + 2 ) 1 2 ( n + 1 ) ( n + 2 ) n ( n − 1 ) = 7 6
After cross-multiplying and transposing, we get:
2 ( n + 1 ) ( n + 2 ) n ( n − 1 ) = ( n + 1 ) ( n + 2 ) 6
Multiplying both sides by ( n + 1 ) ( n + 2 ) since n is n o n n e g a t i v e ,
2 n ( n − 1 ) = 6
n ( n − 1 ) = 1 2
n 2 − n − 1 2 = 0
( n − 4 ) ( n + 3 ) = 0
Since n is n o n n e g a t i v e ,
n − 4 = 0
Therefore, n = 4