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Question

If 1+3p3,1p4 and 12p2 are probabilities of mutually exclusive events of a random experiment, then the range of p is

A
[13,12]
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B
[14,12]
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C
[13,23]
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D
[13,25]
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Solution

The correct option is A [13,12]
Given probabilities are 1+3p3,1p4 and 12p2
Now,
01+3p3113p2301p413p1012p2112p12

From the above cases, we get
13p12 (1)

Also given that events are mutually exclusive, so
01+3p3+1p4+12p2104+12p+33p+612p1210133p12133p113p133 (2)

From (1) and (2) we get
p[13,12]

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