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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 1/3 and 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?

This question was previously asked in
NDA I 2016 GAT Previous Year Paper (17-Apr-2016)
The correct answer is \(\frac{1}{{18}}\)

Cricket Match Probability: Calculating Points in a Series

This problem involves calculating the probability of a cricket team, Team A, achieving a specific total score (5 points) over a series of 3 one-day matches. We are given the probabilities of Team A winning or drawing a single match, along with the points awarded for each outcome (win, loss, or draw).

Step 1: Determine the Probability of Each Outcome for Team A

We are given:

  • Probability of Team A winning ($P(W)$) = $\frac{1}{3}$
  • Probability of Team A drawing ($P(D)$) = $\frac{1}{6}$

In a cricket match, there are three possible outcomes for Team A: Win, Draw, or Loss. The sum of the probabilities of all possible mutually exclusive outcomes must equal 1.

Let $P(L)$ be the probability of Team A losing. Then:

$\qquad P(W) + P(D) + P(L) = 1$

Substituting the given probabilities:

$\qquad \frac{1}{3} + \frac{1}{6} + P(L) = 1$

To find $P(L)$, we subtract the known probabilities from 1:

$\qquad P(L) = 1 - \left(\frac{1}{3} + \frac{1}{6}\right)$

Find a common denominator for the fractions (which is 6):

$\qquad P(L) = 1 - \left(\frac{2}{6} + \frac{1}{6}\right)$

$\qquad P(L) = 1 - \frac{3}{6}$

$\qquad P(L) = 1 - \frac{1}{2}$

$\qquad P(L) = \frac{1}{2}$

So, the probabilities for Team A in a single match are:

  • Win ($P(W)$) = $\frac{1}{3}$
  • Draw ($P(D)$) = $\frac{1}{6}$
  • Loss ($P(L)$) = $\frac{1}{2}$

Step 2: Understand the Points System

The points awarded for each outcome are:

  • Win: 2 points
  • Draw: 1 point
  • Loss: 0 points

Step 3: Identify Outcome Combinations for 5 Points in 3 Matches

Team A plays 3 matches. We need to find combinations of outcomes (Win, Draw, Loss) over these 3 matches that result in a total score of 5 points. Let's list possible scenarios for 3 matches and their total points:

Match 1 Outcome (Points) Match 2 Outcome (Points) Match 3 Outcome (Points) Total Points
W (2) W (2) W (2) 2+2+2 = 6
W (2) W (2) D (1) 2+2+1 = 5
W (2) W (2) L (0) 2+2+0 = 4
W (2) D (1) D (1) 2+1+1 = 4
W (2) D (1) L (0) 2+1+0 = 3
W (2) L (0) L (0) 2+0+0 = 2
D (1) D (1) D (1) 1+1+1 = 3
D (1) D (1) L (0) 1+1+0 = 2
D (1) L (0) L (0) 1+0+0 = 1
L (0) L (0) L (0) 0+0+0 = 0

By examining the possible combinations, we see that the only way for Team A to score exactly 5 points in 3 matches is by winning two matches and drawing one match (2 Wins, 1 Draw). The order of these outcomes can vary.

Step 4: Calculate the Probability of Achieving 2 Wins and 1 Draw

The possible sequences of outcomes over 3 matches that result in 2 Wins and 1 Draw are:

  • Win, Win, Draw (WWD)
  • Win, Draw, Win (WDW)
  • Draw, Win, Win (DWW)

Since the outcome of each match is independent of the others, the probability of a specific sequence is the product of the probabilities of the individual outcomes in that sequence.

  • Probability of WWD = $P(W) \times P(W) \times P(D) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{6} = \frac{1 \times 1 \times 1}{3 \times 3 \times 6} = \frac{1}{54}$
  • Probability of WDW = $P(W) \times P(D) \times P(W) = \frac{1}{3} \times \frac{1}{6} \times \frac{1}{3} = \frac{1 \times 1 \times 1}{3 \times 6 \times 3} = \frac{1}{54}$
  • Probability of DWW = $P(D) \times P(W) \times P(W) = \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{1 \times 1 \times 1}{6 \times 3 \times 3} = \frac{1}{54}$

The event "Team A scores 5 points" is the union of these mutually exclusive sequences (WWD, WDW, DWW). Therefore, the total probability is the sum of the probabilities of these sequences.

$\qquad P(\text{5 points}) = P(\text{WWD}) + P(\text{WDW}) + P(\text{DWW})$

$\qquad P(\text{5 points}) = \frac{1}{54} + \frac{1}{54} + \frac{1}{54}$

$\qquad P(\text{5 points}) = \frac{1 + 1 + 1}{54}$

$\qquad P(\text{5 points}) = \frac{3}{54}$

Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 3:

$\qquad P(\text{5 points}) = \frac{3 \div 3}{54 \div 3} = \frac{1}{18}$

Thus, the probability that Team A will score 5 points in the series is $\frac{1}{18}$.

Revision Table: Cricket Match Probability Concepts

Concept Description Application in Problem
Probability Sum Sum of probabilities of all possible outcomes for an event is 1. $P(W) + P(D) + P(L) = 1$ was used to find $P(L)$.
Independent Events The outcome of one match does not affect the outcome of others. Allowed multiplication of probabilities for sequences like WWD.
Combinations/Permutations Different orders of outcomes (e.g., WWD, WDW, DWW) are distinct sequences. Identified all possible sequences leading to 2 Wins and 1 Draw.
Mutually Exclusive Events Sequences like WWD and WDW cannot happen at the same time. Allowed adding probabilities of different sequences to get total probability.

Additional Information: Probability in Sports Series

Calculating probabilities in sports series often involves understanding independent events and combinations. Each match is usually treated as an independent trial with a fixed set of outcomes (win, loss, draw) and associated probabilities. To find the probability of a specific overall result in a series (like winning a certain number of games, or achieving a specific point total), you typically follow these steps:

  1. Identify the probabilities of each outcome for a single event (one match).
  2. Determine all possible sequences of outcomes over the series that satisfy the desired overall result (e.g., scoring 5 points).
  3. Calculate the probability of each individual sequence by multiplying the probabilities of the outcomes in that sequence (since they are independent).
  4. Sum the probabilities of all the sequences that contribute to the desired overall result (since these sequences are mutually exclusive).

For series with many matches and complex scoring or win conditions, this process can become more complex and might involve binomial or multinomial probability calculations or even simulations. However, for a short series like 3 matches with a simple point system, listing and summing probabilities of sequences is effective.

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