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Question: Probability that A speaks truth is \(\dfrac{4}{5}\). A coin is tossed. A report that a head appears....

Probability that A speaks truth is 45\dfrac{4}{5}. A coin is tossed. A report that a head appears. The probability that actually there was a head is
a. 45\dfrac{4}{5}
b. 12\dfrac{1}{2}
c. 15\dfrac{1}{5}
d. 25\dfrac{2}{5}

Explanation

Solution

Hint: Here, we will solve the given problem by considering all possibilities for the occurrence of an event using the concepts of probability.

Complete step-by-step answer:
Let,
Probability of truth= P(Tr)P(Tr)
Probability of lie= P(F)P(F)
Probability of getting a tail= P(T)P(T)
Probability of getting a head= P(H)P(H)
Now it is given in the question that,
P(Tr)=45P(Tr) = \dfrac{4}{5}
P(F)=145=15P(F) = 1 - \dfrac{4}{5} = \dfrac{1}{5}
Two possibilities can arise:
i) Head has actually occurred.
ii) Head has not occurred but A has lied.
We have to find the probability of heads actually occurred, therefore,
P(TrH)=P(Tr).P(HTr)P(Tr).P(HTr)+P(F).P(TF)P\left( {\dfrac{{Tr}}{H}} \right) = \dfrac{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right)}}{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right) + P(F).P\left( {\dfrac{T}{F}} \right)}}
P(TrH)=45×1245×12+15×12 P(TrH)=45  P\left( {\dfrac{{Tr}}{H}} \right) = \dfrac{{\dfrac{4}{5} \times \dfrac{1}{2}}}{{\dfrac{4}{5} \times \dfrac{1}{2} + \dfrac{1}{5} \times \dfrac{1}{2}}} \\\ P\left( {\dfrac{{Tr}}{H}} \right) = \dfrac{4}{5} \\\
Option a) is correct.

Note: We have taken into consideration all the possibilities carefully, because if any of the possibilities is missed, we will not get the right answer.