If for two events A and B, Pr(A)= \frac{1}{5} , and Pr(B)= \frac{3}{5} , and Pr(A \cup B)= \frac{7}{10} , then Pr(A ∩ B) is equal to:

Question

If for two events A and B, Pr(A)= \frac{1}{5} , and Pr(B)= \frac{3}{5} , and Pr(A \cup B)= \frac{7}{10} , then Pr(A ∩ B) is equal to:

BlackTom AI · Accepted Answer

To find Pr(A ∩ B) we can use the inclusion–exclusion principle: Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).
Rearranging gives Pr(A ∩ B) = Pr(A) + Pr(B) − Pr(A ∪ B).
Plugging in the given values: Pr(A ∩ B) = (1/5) + (3/5) − (7/10).
Compute (1/5) + (3/5) = 4/5 = 8/10. Then 8/10 − 7/10 = 1/10.
Therefore, the intersection probability is 1/10. This matches the option listed as 'a. 1/10'.

Why this makes sense: the intersection cannot be larger than either Pr(A) or Pr(B), and the union cannot exceed 1. The calculated 1/10 sits within those bounds and is consistent with all provided marginals.

Why other hypothetical options (if present) would be wrong: if you tried to add Pr(A ∩ B) to Pr(A ∪ B) and compare to Pr(A) + Pr(B), you’d violate the inclusion–exclusion relationship. If you assumed independence without the given union, you’d compute Pr(A)Pr(B) = (1/5)(3/5) = 3/25 = 0.12, which is 0.12, not aligning with the provided union probability 7/10; hence independence is not implied by the data.

MUF0092 Mathematics Unit 2-SEM 1-2025 7.15 Probability and discrete probability distributions quiz

If for two events A and B, Pr(A)= \frac{1}{5} , and Pr(B)= \frac{3}{5} , and Pr(A \cup B)= \frac{7}{10} , then Pr(A ∩ B) is equal to:

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