Question

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

Answer 1

Hii bro

First of all thanks for posting a beautiful question

Solution :-

Here the random experiment is :- The stating of fact by A and B

The fact by A and B

let E = The event of A speaking the truth

and F = the event of B speaking the truth

Then , P(E) = 60/100 = 3/5 and P(F) = 90/100 = 9/10

Required probability , P(A and B contradict each other )

= P(E not F or Not E F) = P(E not F ) + P(not E F)

= P(E) * P(not F) + P(not E) * P(F)

=> P(E)* [1-P(E) ] + [1-p(E)]*P(F)

= 3/5(2-9/10)+(2-3/5)*9/10 = 21/50

•°• A and B are likely to contradict each other in 42% cases .

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Hope it's helpful

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Answer 2

$\huge{\underline{\blue{\mathfrak{Solution:-}}}$

Let the probability that A and B speak truth be P(A) and P(B) respectively.

$\sf {Therefore,\ P(A) = \frac{60}{100} = \frac{3}{5}\ and\ P(B) = \frac{90}{100}}=\frac{9}{10}$

A and B can contradict in stating a fact when one is speaking the truth and other is not speaking the truth.

Case 1: A is speaking the truth and B is not speaking the truth.

$\sf {Required\ probability = P(A)\times (1-P(B))} = \frac{3}{5}\times (1-\frac{9}{10} = \frac{3}{5})$

Case 2: A is not speaking the truth and B is speaking the truth.

$\sf {Required\ probability = (1 -P(A))\times P(B) = (1-\frac{3}{5})\times 9 = \frac{9}{25})}$

Percentage of cases in which they are likely to contradict in stating the same fact =

$\sf {(\frac{3}{50}+\frac{9}{25})\times 100\ percent = (\frac{3+18}{50})\times 100\ percent = 42\ percent }$