Let's solve each part of the problem step by step.

(a) Probability that the student is a woman

The total number of students is $49$, and the number of women is $38$.

The probability that the student is a woman is given by: $P(\text{Woman}) = \frac{\text{Number of women}}{\text{Total number of students}} = \frac{38}{49}$

(b) Probability that the student received an A

The total number of students who received an A is the sum of the men and women who received an A: $\text{Number of students who received an A} = 2 (\text{men}) + 3 (\text{women}) = 5$

The probability that the student received an A is: $P(\text{A}) = \frac{\text{Number of students who received an A}}{\text{Total number of students}} = \frac{5}{49}$

(c) Probability that the student is a woman or received an A

To find this probability, we need to use the formula for the union of two events: $P(\text{Woman or A}) = P(\text{Woman}) + P(\text{A}) - P(\text{Woman and A})$

First, we calculate $P(\text{Woman and A})$, which is the probability that the student is a woman who received an A: $P(\text{Woman and A}) = \frac{\text{Number of women who received an A}}{\text{Total number of students}} = \frac{3}{49}$

Now, substituting the values: $P(\text{Woman or A}) = \frac{38}{49} + \frac{5}{49} - \frac{3}{49} = \frac{40}{49}$

(d) Probability that the student did not receive an A

The probability that the student did not receive an A is the complement of the probability that the student did receive an A: $P(\text{Not A}) = 1 - P(\text{A}) = 1 - \frac{5}{49} = \frac{44}{49}$

Summary of Probabilities

• (a) Probability that the student is a woman: $\frac{38}{49}$