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# A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 13, then the correct option(s) with the possible values of n1 and n2 is (are)n1=4 and n2=6n1=2 and n2=3n1=3 and n2=6n1=10 and n2=20

A
n1=4 and n2=6
B
n1=3 and n2=6
C
n1=2 and n2=3
D
n1=10 and n2=20
Solution
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#### 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.Let E3 be the event that transferred ball is red and E4 be the event that transferred ball is black.R be the event of selecting a Red ball.P(R)=P(E3).P(R/E3)+P(E4).P(R/E4)⇒P(R)=13=n1n1+n2.(n1−1n1+n2−1)+n2n1+n2.(n1n1+n2−1)=n1n1+n2only options C and D satisfies the above relation.

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