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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?

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0.40

0.58

0.42

None of these

answer is C.

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Detailed Solution

Suppose,
Event for A speaking truth =

E

Event for B speaking truth

= F

Given,
A speaks truth = 60% cases
B speaks truth =90% cases
For whole part total cases = 100
So, probability of A speaking truth,

P (E) = \frac{Favorable outcome}{total outcome}

P (E) = \frac{60}{100}

= \frac{6}{10}

Also, probability of B speaking truth,

P (F) = \frac{Favorable outcome}{total outcome}

P (F) = \frac{90}{100}

= \frac{9}{10}

Now, the probability of A and B contradicting one other

P (EF or F^{¯} E)

can be given as,

P (E F^{\bar{}}) + P (E^{\bar{}} F) = P (E) P (F^{\bar{}}) + P (E^{\bar{}}) P (F)

= P (E) (1 - P (F)) + (1 - P (E)) P (F)

= \frac{6}{10} (1 - \frac{9}{10}) + (1 - \frac{6}{10}) \frac{9}{10}

= \frac{6}{10} \times \frac{1}{10} + \frac{4}{10} \times \frac{9}{10}

= \frac{42}{100}

= 0.42

As a result, A and B are expected to be in conflict in 42% of cases, or 0.42.
Hence, correct option is 3.

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