To solve the problem, we will use the principles of probability and the concept of the union of events.
Let:
- \( A \): the event that a student got an A in Statistics
- \( B \): the event that a student got an A in Psychology
From the information provided:
- Total students (\( n \)) = 700
- \( n(A) \) = Number of students who got an A in Statistics = 76
- \( n(B) \) = Number of students who got an A in Psychology = 72
- \( n(A \cap B) \) = Number of students who got an A in both Statistics and Psychology = 45
(a) Probability that a randomly chosen student got an A in Statistics or Psychology or both:
We use the formula for the union of two sets:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
First, we need to find the individual probabilities:
\[ P(A) = \frac{n(A)}{n} = \frac{76}{700} \] \[ P(B) = \frac{n(B)}{n} = \frac{72}{700} \] \[ P(A \cap B) = \frac{n(A \cap B)}{n} = \frac{45}{700} \]
Now we can substitute into the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{76}{700} + \frac{72}{700} - \frac{45}{700} \]
Calculating the probabilities:
\[ P(A \cup B) = \frac{76 + 72 - 45}{700} = \frac{103}{700} \]
Finally, to get the probability:
\[ P(A \cup B) = \frac{103}{700} \approx 0.1471 \]
(b) Probability that a randomly chosen student did not get an A in Psychology:
The event that a student did not get an A in Psychology is the complement of the event \( B \):
\[ P(B') = 1 - P(B) \]
We already calculated \( P(B) \):
\[ P(B) = \frac{72}{700} \]
Thus:
\[ P(B') = 1 - \frac{72}{700} = \frac{700 - 72}{700} = \frac{628}{700} \]
Calculating this probability gives:
\[ P(B') = \frac{628}{700} \approx 0.8971 \]
Summary of Answers:
(a) The probability that a randomly chosen student got an A in Statistics or Psychology or both is approximately 0.1471.
(b) The probability that a randomly chosen student did not get an A in Psychology is approximately 0.8971.