Question

In a series of 3 one-day cricket matches between teams A and B of a college, the probability of team A winning or drawing are $\dfrac{1}{3}$ and $\dfrac{1}{6}$ respectively. If a win, loss or draw gives 2, 0 and 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

This problem deals with probability. The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes of an event. Here the concept of independent events is used. The independent events are those events which are not affected by the previous events. That is, we say two events are independent if knowing one event occurred doesn’t change the probability of the other event.
Let A and B are two independent events, then it is given by:
$\Rightarrow P\left( {AB} \right) = P\left( A \right)P\left( B \right)$

Complete step-by-step solution:
Given that one-day cricket matches are happening in 3 serial days.
These one-day cricket matches are between two teams A and B from a college.
Let us denote win for team A as ‘’
Let us denote draw for team A as ‘$D$’
Given that the probability of team A winning is $\dfrac{1}{3}$
$\Rightarrow P\left( W \right) = \dfrac{1}{3}$
The probability of team A drawing = $\dfrac{1}{6}$
$\Rightarrow P\left( D \right) = \dfrac{1}{6}$
Also given that there are certain points awarded to win, loss and draw.
The number of points given if team A wins = $2$
The number of points given if team A draws = $1$
The number of points given if team A loses = $0$
We have to find the probability in the series that team A scores 5 points.
So in order to score 5 points for team A, they should win 2 matches and draw 1 match, and this can be in any order.
So there has to be 2 wins and 1 draw match for team A to win, the possible 3 outcomes are given by:
$\Rightarrow WWD + WDW + DWW$
The probability to score 5 points is given by:
$\Rightarrow P\left( {WWD} \right) + P\left( {WDW} \right) + P\left( {DWW} \right)$
As the events win and draw are independent events, hence splitting the probability of win and draw and multiply them, as given below:
$\Rightarrow P\left( W \right)P\left( W \right)P\left( D \right) + P\left( W \right)P\left( D \right)P\left( W \right) + P\left( D \right)P\left( W \right)P\left( W \right)$
$\Rightarrow \left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{6}} \right) + \left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{6}} \right)\left( {\dfrac{1}{3}} \right) + \left( {\dfrac{1}{6}} \right)\left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{3}} \right)$
Substituting the values of $P\left( W \right) = \dfrac{1}{3}$and $P\left( D \right) = \dfrac{1}{6}$, as shown above.
$\Rightarrow \dfrac{1}{{54}} + \dfrac{1}{{54}} + \dfrac{1}{{54}} = \dfrac{3}{{54}}$
$\Rightarrow \dfrac{1}{{18}}$
$\therefore P\left( {WWD} \right) + P\left( {WDW} \right) + P\left( {DWW} \right) = \dfrac{1}{{18}}$
The probability that team A will score 5 points in the series is $\dfrac{1}{{18}}$.

Option D is the correct answer.

Note: Here while solving the problem we have to be careful when considering the probability of the events. Here we are asked to consider only the probability of team A scoring 5 points. Here it is only possible if the team scores 2 wins at it scores 2 points for each win, hence 2 wins scores 4 points for the team, and another point is needed to score 5 points, hence draw is considered.