
Probability that A speaks truth is $\dfrac{4}{5}$. A coin is tossed. A report that a head appears. The probability that actually there was a head is
a. $\dfrac{4}{5}$
b. $\dfrac{1}{2}$
c. $\dfrac{1}{5}$
d. $\dfrac{2}{5}$
Answer
208.5k+ views
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= $P(Tr)$
Probability of lie= $P(F)$
Probability of getting a tail= $P(T)$
Probability of getting a head= $P(H)$
Now it is given in the question that,
$P(Tr) = \dfrac{4}{5}$
$P(F) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$
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,
$P\left( {\dfrac{{Tr}}{H}} \right) = \dfrac{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right)}}{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right) + P(F).P\left( {\dfrac{T}{F}} \right)}}$
$
P\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.
Complete step-by-step answer:
Let,
Probability of truth= $P(Tr)$
Probability of lie= $P(F)$
Probability of getting a tail= $P(T)$
Probability of getting a head= $P(H)$
Now it is given in the question that,
$P(Tr) = \dfrac{4}{5}$
$P(F) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$
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,
$P\left( {\dfrac{{Tr}}{H}} \right) = \dfrac{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right)}}{{P(Tr).P\left( {\dfrac{H}{{Tr}}} \right) + P(F).P\left( {\dfrac{T}{F}} \right)}}$
$
P\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.
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