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 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).
We are given:
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:
The points awarded for each outcome are:
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.
The possible sequences of outcomes over 3 matches that result in 2 Wins and 1 Draw are:
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.
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 $\frac{1}{18}$.
| 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. |
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:
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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