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Probability

Long Answer Type

871.

The probability of A solving a problem is $3 over 7$ and that of B solving it is $1 third.$ What is the probability that
(i) at least one of them will solve the problem?
(ii) Only one of them will solve the problem?

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Short Answer Type

872. 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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873. A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?

Let P(A), P(B) be the probability of A and B are speaking the truths. Then
$straight P left parenthesis straight A right parenthesis space equals space 75 over 100 space equals 3 over 4 comma space space space straight P left parenthesis straight B right parenthesis space equals 80 over 100 space equals 4 over 5$
$therefore space space space space space straight P left parenthesis straight A with bar on top right parenthesis space equals space straight P left parenthesis straight A space tells space straight a space lie right parenthesis space equals space 1 space space minus straight P left parenthesis straight A right parenthesis space equals space 1 minus space 3 over 4 space equals 1 fourth space space space space space space space space space straight P left parenthesis straight B with bar on top right parenthesis space equals space straight P left parenthesis straight B space tells space straight a space lie right parenthesis space equals space 1 minus straight P left parenthesis straight B right parenthesis space equals space 1 minus 4 over 5 space equals 1 fifth$
Now, P(A and B contradict) = $straight P left parenthesis straight A right parenthesis space straight P left parenthesis straight B with bar on top right parenthesis space plus space straight P left parenthesis straight A with bar on top right parenthesis space straight P left parenthesis straight B right parenthesis$
$space equals space 3 over 4 cross times 1 fifth plus 4 over 5 cross times 1 fourth space equals space 3 over 20 plus 1 fifth space equals 7 over 20 equals space 35 percent sign$

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874. The probability of hitting a target by three marksmen are

1 half comma space 1 third space and space 1 fourth

respectively. Find the probability that one and only one of them will hit the target when they fire simultaneously.

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875. An anti-aircraft gun can take a minimum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?

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