Question

An event X can take place in conjunction with any one of the mutually exclusive and exhaustive events A, B and C. If A, B,C are equiprobable and the probability of X is \[\dfrac{5}{{12}}\]and the probability of X taking place when A has happened is \[\dfrac{3}{8}\]while it is \[\dfrac{1}{4}\]when B has taken place, then the probability of X taking place on conjunction with C is
A. \[\dfrac{5}{8}\]
B. \[\dfrac{3}{8}\]
C. \[\dfrac{5}{{24}}\]
D. none of these.

Answer 1

Hint: Here, it is given the conjunction of A, conjunction of B, and conjunction of C with the probability of X. It means they are denoted as \[P(\dfrac{X}{A})\], \[P(\dfrac{X}{B})\] and \[P(\dfrac{X}{C})\], so the probability is the number of outcome of an event from a number of trials. In this solution, equate all probabilities with the probability of the X.

Complete step-by-step answer:
The probability of X when A, B, and C are equiprobable is \[P(X) = \dfrac{5}{{12}}\]
The probability of X when the A has happened is \[P(\dfrac{X}{A}) = \dfrac{3}{8}\]
The probability of X when the B has happened is \[P(\dfrac{X}{B}) = \dfrac{1}{4}\]
It can be said that the probability of A, B and C are equal, then it can be said that,
\[P(A) = P(B) = P(C) = \dfrac{1}{3}\]
The equation to find the probability of X when the C has happened is
\[P(X) = P(A) \times P(\dfrac{X}{A}) + P(B) \times P(\dfrac{X}{B}) + P(C) \times P(\dfrac{X}{C})\]
Substituting the values in the above equation, then
\[\begin{array}
P(X) = P(A) \times P(\dfrac{X}{A}) + P(B) \times P(\dfrac{X}{B}) + P(C) \times P(\dfrac{X}{C})\\
\dfrac{5}{{12}} = \dfrac{1}{3} \times \dfrac{3}{8} + \dfrac{1}{3} \times \dfrac{1}{4} + \dfrac{1}{3} \times P(\dfrac{X}{C})
\end{array}\]
On solving further,
\[\begin{array}
\dfrac{5}{{12}} = \dfrac{3}{{24}} + \dfrac{1}{{12}} + \dfrac{1}{3} \times P(\dfrac{X}{C})\\
\dfrac{5}{{12}} = \dfrac{5}{{24}} + \dfrac{1}{3} \times P(\dfrac{X}{C})\\
P(\dfrac{X}{C}) = 3\left( {\dfrac{5}{{12}} - \dfrac{5}{{24}}} \right)\\
P(\dfrac{X}{C}) = 3 \times \dfrac{5}{{24}}\\
P(\dfrac{X}{C}) = \dfrac{5}{8}
\end{array}\]
Therefore, the probability of X taking place in conjunction with C is \[\dfrac{5}{8}\], it means that the option (A) is correct.
So, the correct answer is “Option A”.

Note: In this solution, we may be confused that what term should find out, so we should be clear about that term and it is should remember the statement given in the question that “A, B and C are equiprobable” which means the probability of all the terms are equal, so we can assume it as the \[\dfrac{1}{3}\]because the number of terms are 3 and when we draw anything one at a time then the probability will be\[\dfrac{1}{3}\].