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Question

Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II.

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 13, then the correct option(s) with the possible values of n1,n2,n3 and n4 is(are)

A
n1=3,n2=3,n3=5,n4=15
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B
n1=3,n2=6,n3=10,n4=50
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C
n1=8,n2=6,n3=5,n4=20
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D
n1=6,n2=12,n3=5,n4=20
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Solution

The correct options are
A n1=3,n2=3,n3=5,n4=15
B n1=3,n2=6,n3=10,n4=50
Box I:
Red balls n1 and black balls n2
Box II:
Red balls n3 and black balls n4
Probability of drawing a red ball =P(R)
P(R)=12n1n1+n2+12n3n3+n4

Now probability that box II was selected,
P(II/R)=P(IIR)P(R)12n3n3+n4P(R)=13n3n3+n4n1n1+n2+n3n3+n4=13n1n1+n2n3n3+n4+1=32n3n3+n4=n1n1+n2

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