To solve the problems, we can use the principles of set theory and probability.
Let:
- \( P(S) \) be the probability that a student got an A in Statistics.
- \( P(P) \) be the probability that a student got an A in Psychology.
- \( P(S \cap P) \) be the probability that a student got an A in both Statistics and Psychology.
Given:
- Total number of students, \( n = 300 \)
- Number of students who got an A in Statistics, \( n(S) = 90 \)
- Number of students who got an A in Psychology, \( n(P) = 78 \)
- Number of students who got an A in both Statistics and Psychology, \( n(S \cap P) = 40 \)
(a) Probability that a student got an A in Statistics or Psychology or both
We want to find \( P(S \cup P) \), which can be calculated using the formula:
\[ P(S \cup P) = P(S) + P(P) - P(S \cap P) \]
Calculating each probability:
- \( P(S) = \frac{n(S)}{n} = \frac{90}{300} = 0.3 \)
- \( P(P) = \frac{n(P)}{n} = \frac{78}{300} = 0.26 \)
- \( P(S \cap P) = \frac{n(S \cap P)}{n} = \frac{40}{300} = 0.1333 \)
Now substituting these values into the formula:
\[ P(S \cup P) = 0.3 + 0.26 - 0.1333 = 0.4267 \]
So, rounding to four decimal places, we get:
\[ P(S \cup P) \approx 0.4267 \]
(b) Probability that a student did not get an A in Psychology
To find the probability that a student did not get an A in Psychology, we first find \( P(P') \), where \( P' \) represents the complement of event \( P \):
\[ P(P') = 1 - P(P) \]
We already calculated \( P(P) \):
\[ P(P') = 1 - 0.26 = 0.74 \]
Thus, rounding to four decimal places, we obtain:
\[ P(P') \approx 0.7400 \]
Summary of Results:
(a) The probability that a randomly chosen student got an A in statistics or psychology or both is 0.4267.
(b) The probability that a randomly chosen student did not get an A in psychology is 0.7400.
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