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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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The probability that a student passes an English test is 32 and the probability that he passes both the English and a Science test is 41. The probability that he passes at least one test is 1211. What is the probability that he passes the Science test?
1/2
1/3
1/4
2/5
1/2
Use the direct relation: P(S) = P(E ∪ S) + P(E ∩ S) - P(E). Simply calculate (11/12 + 1/4) - 2/3 = (14/12) - 8/12 = 6/12 = 1/2.
Probability of passing English P(E) = 2/3, Probability of passing both P(E ∩ S) = 1/4, Probability of passing at least one P(E ∪ S) = 11/12.
P(E∪S)=P(E)+P(S)−P(E∩S)
Use the direct relation: P(S) = P(E ∪ S) + P(E ∩ S) - P(E). Simply calculate (11/12 + 1/4) - 2/3 = (14/12) - 8/12 = 6/12 = 1/2.
Students often confuse the union P(E ∪ S) with the intersection P(E ∩ S) or incorrectly assume the tests are independent events.
State the Addition Theorem of Probability
To find the probability of at least one event occurring, we use the formula for the union of two sets.
P(E∪S)=P(E)+P(S)−P(E∩S)
Substitute given values
Plug the known values into the equation: P(E)=2/3, P(E∩S)=1/4, and P(E∪S)=11/12.
1211=32+P(S)−41
Solve for P(S)
Isolate P(S) by moving constants to one side: P(S)=1211−32+41. Find the common denominator, which is 12.
P(S)=1211−8+3=126=21
A is correct because the calculation P(S)=21 matches the given option.
This is equivalent to the Principle of Inclusion-Exclusion for two sets: ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣, which is frequently used in counting problems and Venn diagram questions.