If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then, the probability that either A or B fails is:
0.38
This question asks for the probability that either student A or student B fails an examination. We are given the individual probabilities of failure for each student.
Let's define the events:
We are given the following probabilities:
The question asks for the probability that "either A or B fails". Based on the provided answer options, this phrase is interpreted as the probability that *exactly one* of the students fails the examination. This can occur in two distinct scenarios:
To calculate these, we first need the probabilities of each student passing:
We assume that the events of A failing and B failing are independent. This is a common assumption in such problems unless stated otherwise.
Now, we calculate the probability for each scenario:
Using the independence assumption, the probability of this scenario is:
$$ P(\text{A fails AND B passes}) = P(A_{fail}) \times P(B_{pass}) $$
$$ P(\text{A fails AND B passes}) = 0.2 \times 0.7 = 0.14 $$
Similarly, the probability of this scenario is:
$$ P(\text{A passes AND B fails}) = P(A_{pass}) \times P(B_{fail}) $$
$$ P(\text{A passes AND B fails}) = 0.8 \times 0.3 = 0.24 $$
Since Scenario 1 and Scenario 2 are mutually exclusive events (they cannot happen at the same time), the total probability that *either* A or B fails (meaning exactly one fails) is the sum of their probabilities:
$$ P(\text{Exactly one fails}) = P(\text{A fails AND B passes}) + P(\text{A passes AND B fails}) $$
$$ P(\text{Exactly one fails}) = 0.14 + 0.24 $$
$$ P(\text{Exactly one fails}) = 0.38 $$
Therefore, the probability that either A or B fails the examination is 0.38.
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