Question

In a series of \[3\] one-day cricket matches between teams \[A{\text{ }}and{\text{ }}B\]of a college, the probability of team A winning or drawing is \[\dfrac{1}{3}\] and \[\dfrac{1}{6}\] respectively. If a win, lose or draw gives \[2,{\text{ }}0{\text{ }}and{\text{ }}1\] point respectively, then what is the probability that team A will score \[5\] points in the series?
A.\[\dfrac{{17}}{{18}}\]
B.\[\dfrac{{11}}{{12}}\]
C.\[\dfrac{1}{{12}}\]
D.\[\dfrac{1}{{18}}\]

Answer 1

Hint: First, we should know the basic formula of probability which is given as-
Probability of an event $P\left( E \right) = \dfrac{{{\text{Number of outcome}}}}{{{\text{total sample space}}}}$
In this type of problem we use the possibility of occurrence of an event and then all probabilities are added.

Complete step-by-step answer:
Given in the question,
Probability of team A winning\[ = \dfrac{1}{3}\]
Probability of team A drawing\[ = \dfrac{1}{6}\]
Point for winning $ = 2$
Points for losing$ = 0$
Points for draw$ = 1$
Let, Probability of team A winning and drawing be $X$and $Y$.
To score \[5\] points team A has to win \[2\]times and draw \[1\] time.
So the possible outcomes are $\left( {XXY,XYX,YXX} \right)$.
So, probability to score \[5\] points by team $A$\[ = \]
$ = P(X)P(X)P(Y) + P(X)P(Y)P(X) + P(Y)P(X)P(X)$
\[ = \left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{6}} \right) + \left( {\dfrac{1}{3} \times \dfrac{1}{6} \times \dfrac{1}{3}} \right) + \left( {\dfrac{1}{6} \times \dfrac{1}{3} \times \dfrac{1}{3}} \right)\]
It can be further simplified,
\[\begin{gathered}
 = \dfrac{1}{{54}} + \dfrac{1}{{54}} + \dfrac{1}{{54}} \\
 = \dfrac{3}{{54}} = \dfrac{1}{{18}} \\
\end{gathered} \]
Therefore, the probability of team $A$ scoring \[5\]points is \[\dfrac{1}{{18}}\]
So, the correct answer is “Option C”.

Note: Above in the solution, the probability of occurrence of an event is multiplied and then all possible events are added to obtain the probability of team A scoring \[5\] points. Avoid calculation mistakes during solving the problem.