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- n1=4 and n2=6
- n1=2 and n2=3
- n1=3 and n2=6
- n1=10 and n2=20

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Solution

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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+n2

only options C and D satisfies the above relation.

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