To solve this problem, we can use the principles of probability. Here's a step-by-step breakdown:

Step 1: Define the variables

• Total students (T): 19
• Female students (F): 6
• Students with an A (A): 10
• Male students without an A (M' ∩ A'): 7

Step 2: Determine the number of male students

Since the total number of students is 19 and there are 6 female students, the number of male students $M$ is: $M = T - F = 19 - 6 = 13$

Step 3: Determine the number of male students with an A

We know that the number of male students who do not have an A is 7. Since there are 13 male students in total, the number of male students with an A $M \cap A$ is: $M \cap A = M - (M' \cap A') = 13 - 7 = 6$

Step 4: Calculate the probability

We need to find the probability that a student who has an A is a male. This can be represented as: $P(M \mid A) = \frac{P(M \cap A)}{P(A)}$ Where:

• $P(M \cap A)$ is the number of male students with an A, which is 6.
• $P(A)$ is the total number of students with an A, which is 10.

Substituting the values, we get: $P(M \mid A) = \frac{6}{10} = \frac{3}{5} = 0.6$

Final Answer

The probability that a student who has an A is a male is $\boxed{0.6}$ or 60%.

Would you like more details or have any questions?

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Here are some related questions:

1. What is the probability that a student selected at random is a female with an A?
2. What is the probability that a student who is male does not have an A?
3. How many students have neither an A nor are male?
4. What is the probability that a student selected at random is male?
5. How does the addition of one more female student with an A affect the probability?

Tip: When dealing with conditional probabilities, clearly identify the events and use the conditional probability formula $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.