Answer:
The probability that a student is taking both math and computer science
P(M∩CS) = 0.10
Step-by-step explanation:
Explanation:-
Given data the probability that a student is taking math is 23%
Let 'M' be the event of a student is taking math
P(M) = 0.23
Let 'CS' be the event of a student is taking computer science
Given data the probability that a student is taking computer science is 45%
P(CS) = 0.45
Given the probability that a student is taking math or computer science is 58%.
P(M U CS) = 0.58
Addition theorem on probability
If S is a sample size , and E₁ and E₂ be the events in S then
P(E₁ or E₂ ) = P(E₁) + P(E₂) - P(E₁ and E₂)
or
P(E₁ ∪E₂ ) = P(E₁) + P(E₂) - P(E₁ ∩ E₂)
Now
P(M∪ CS ) = P(M) + P(CS) - P(M∩CS)
0.58 = 0.23 + 0.45 - P(M∩CS)
P(M∩CS) = 0.68 -0.58
P(M∩CS) = 0.10
Conclusion:-
The probability that a student is taking both math and computer science
P(M∩CS) = 0.10
