An international team has two boxers picked for an international sport event. What is the probability that both the boxers are men given that at least one of them is a man?
Let's break down this probability problem step-by-step to find the likelihood that both boxers are men, given that we already know at least one of them is a man. This is a classic example of conditional probability.
Boxers: Understanding the Sample Space
First, we need to list all possible gender combinations for two boxers picked for an international sport event. Let 'M' represent a male boxer and 'W' represent a female boxer. The possible outcomes for the two boxers are:
| Boxer 1 | Boxer 2 | Outcome |
|---|---|---|
| Male (M) | Male (M) | (M, M) |
| Male (M) | Female (W) | (M, W) |
| Female (W) | Male (M) | (W, M) |
| Female (W) | Female (W) | (W, W) |
The total number of possible outcomes in our sample space (S) is 4. So, \(n(S) = 4\).
Defining the Events
To calculate the conditional probability, we need to define two specific events:
- Event A: Both boxers are men.
- This event consists of only one outcome: (M, M).
- So, \(A = \{ (M, M) \}\) and \(n(A) = 1\).
- The probability of Event A is \(P(A) = \frac{n(A)}{n(S)} = \frac{1}{4}\).
- Event B: At least one of the boxers is a man.
- "At least one man" means we have one man and one woman, or two men.
- This event includes the outcomes: (M, M), (M, W), (W, M).
- So, \(B = \{ (M, M), (M, W), (W, M) \}\) and \(n(B) = 3\).
- The probability of Event B is \(P(B) = \frac{n(B)}{n(S)} = \frac{3}{4}\).
- Event A \(\cap\) B: Both boxers are men AND at least one of them is a man.
- If both boxers are men (Event A), it automatically means at least one of them is a man (Event B). Therefore, the intersection of A and B is simply Event A itself.
- So, \(A \cap B = \{ (M, M) \}\) and \(n(A \cap B) = 1\).
- The probability of Event A \(\cap\) B is \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{4}\).
Conditional Probability Calculation
We are asked to find the probability that both boxers are men, given that at least one of them is a man. This is denoted as \(P(A|B)\), which is the conditional probability of Event A occurring given that Event B has already occurred.
The formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Now, let's plug in the probabilities we calculated:
- \(P(A \cap B) = \frac{1}{4}\)
- \(P(B) = \frac{3}{4}\)
So, the calculation becomes:
\[ P(A|B) = \frac{\frac{1}{4}}{\frac{3}{4}} \]
To simplify, we can multiply the numerator by the reciprocal of the denominator:
\[ P(A|B) = \frac{1}{4} \times \frac{4}{3} \]
\[ P(A|B) = \frac{1}{3} \]
Conclusion
The probability that both boxers are men, given that at least one of them is a man, is \(\frac{1}{3}\).
Important Questions from Probability of Random Experiments
A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
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From two well shuffled pack of cards what is the probability of getting one Jack from the first one and a King from the second?
If A is an event of getting 13 by throwing two unbiased six-faced dice, then A is called
One summer, Peter visits 4 villages (A, B, C and D) in a random order. Find the probability that he visits (i) A before B (ii) A before B and B before C respectively?
