Engineering
Mathematics
Conditional Probability
Question

If 1+4p4, 1p2, 12p2 are probabilities of three mutually exclusive events, then

13p12

13p23

16p12

None of these

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Solution
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As 1+4p4, 1p2, 12p2 are probabilities of three mutually exclusive events, we must have

01+4p41, 01p21, 012p21 and 01+4p4+1p2+12p21
14p34, 1p1, 12p12 and 12p52
max14, 1, 12,12<pmin34, 1, 12, 52
12p12p=12
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