Answer:
See the Venn diagram corresponding to this question on the picture attached.
Explanation:
Call F the event playing football, P(F) the probability of playing football, S the event of playing baseball, P(S) the probability of playing baseball, K the event of playing basketball, and P(K) the probability of playing basketball.
The probability that a student plays football or baseball is P(F ∪ S)
By the properties of joint probabilities:
- P(F ∪ S) = P(F) + P(S) - P (F∩S)
Each of the terms on the right-hand side of the equation can be calculated from the Venn diagram.
a) Number of total students represented on the Venn diagram:
- 26 + 9 + 10 + 6 + 3 + 7 + 5 = 66
b) Number of students that play football:
Therefore, P(F) = 40 / 66 = 20 / 33
c) Number of students that play baseball:
Therefore, P(S) = 24 / 66 = 4/11
d) Number of students that play both football and baseball:
Therefore, P (F ∩ S) = 8/66 =4/33
Finally:
- P(F ∪ S) = P(F) + P(S) - P (F∩S) = 20/33 + 4/11 - 4/33
- P(F ∪ S) = 16/33 + 4/11 = 16/33 + 12/33 = 28/33
Answer: 28/33
