AkshdeepS wrote:
The chance of rain in Dennyville is 40% on Monday and 80% on Tuesday. What is the probability that it rains in Dennyville on at least one of these two days?
1. 120%
2. 88%
3. 76%
4. 60%
5. 32%
P(at least one) = 1 - P(none)For "at least one" probability questions, it is usually easier to find the probability of NONE and subtract from 1. We have only one case to consider: 0 days of rain.
The opposite of "at least one" is "none." P(at least one) = 1 - P(none)
P(none) = P(no rain on M and no rain on Tu)
P(no rain on M and no rain on Tu):
\((0.60 * 0.20) = (0.12)\)Thus P(at least one day) = 1 - P(0 days):
\((1 - 0.12) = .88 = 88\) %
ANSWER B
ExplanationIf the events are independent (probability or outcome on Monday does not affect probability or outcome on Tuesday), and
"Success"
= it DOES rain on at least one day
= P(A), then
"Failure" (no successes)
= it DOES NOT rain on at least one day
= it rains on 0 days
= P(not A)
The complement rule applies
P(A) + P(not A) = 1
P(A) = 1 - P(not A)
P(at least 1 success) = 1 - P(no successes)
--If "success" contains multiple possibilities, and
-- the problem asks for "at least" or "at most," then
-- try finding the probability that success
does not occur, i.e., try finding the probability of failure. Subtract from 1.
The result is the probability of "at least one."
For an overview of "at least one," see
Probability - At least, on GMAT Club, here.
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