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Question

A speaks truth in 60% cases and B in 90% cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

A
0.42
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B
0.58
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C
0.46
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D
None of these
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Solution

The correct option is A 0.42
Let E= The event of A speaking truth and
F= event of B speaking truth

P(E)=60100=610 and P(F)=90100=910

Probability of A and B contradicting each other P(E¯¯¯¯F or ¯¯¯¯FE)
P(E¯¯¯¯F)+P(¯¯¯¯EF)=P(E)P(¯¯¯¯F)+P(¯¯¯¯E)P(F)
=P(E)(1P(F))+(1P(E))P(F)
=610(1910)+(1610)910
=610×110+410×910=42100
A and B are likely to contradicts each other in 42% cases.

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