Question

p speaks truth in 80% of the cases and q in 90% of the cases. while stating the same fact, what is the chance that they will contradict?

Answer 1

let A be Event that p speaks the truth
and B be Event that q speaks the truth.

P(A) = 80/100 = 4/5
P(B) = 90/100 = 9/10

A' = not A
=>A' is the Event that p tells a lie
B' = not B
=>B' is the Event that q tells a lie

There are only 2 events for A (also for B)
i.e., Either p speaks the truth or p tells a lie.
//y Either q speaks the truth or q tells a lie

Sum of probability is always 1
so,
We have P(A) + P(A') = 1 similarly P(B) + P(B') = 1

°.° P(A') = 1 - (4/5) = 1/5
P(B') = 1 - (9/10) = 1/10

P(p and q contradict each other):
=P[(p speaks the truth and q tells a lie) or (p tells a lie and q speaks the truth)]
=P(A and B') or P(A' and B)
= P(A and B') + P(A' and B)

{we have,
If A and B are independent events, then
P(A and B)= P(A) . P(B). }

Required probability
= P(A).P(B') + P(A').P(B)

=(4/5).(1/10) + (9/10).(1/5)
=(4/50) + (9/50)
=13/50
=[(13/50)*100]%
= 26%
.•. p and q contradict each other in 26% of the cases.



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