The probability that a teacher will give an unannounced test during any class meeting is 1/5.If a student is absent twice,then the probability that the student will miss at least one test is
A.4/5
B.2/5
C.7/75
D.9/25
E.8/25
Not sure what the official answer is. Confused between two of the options. Help please !
Probability problem
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- Anurag@Gurome
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Probability of missing at least one test = Probability of missing test in only one day + Probability of missing test on both the dayauppala wrote:The probability that a teacher will give an unannounced test during any class meeting is 1/5.If a student is absent twice,then the probability that the student will miss at least one test is
Probability of missing test in only one day = (Probability of happening test on one of the day he was absent)*(Probability of not happening test on the other day he was absent) = (1/5)*(1 - 1/5) = (1/5)*(4/5) = 4/25
Now the above can happen in two ways : {test on first day, no test on second day} or {no test on first day, test on second day}
Hence, probability of missing test in only one day = 2*(4/25) = 8/25
And, probability of missing test on both the day = (1/5)*(1/5) = 1/25
Hence, required probability = 8/25 + 1/25 = 9/25
The correct answer is D.
Anurag Mairal, Ph.D., MBA
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Questions with "at least" are often great candidates for using the complement.auppala wrote:The probability that a teacher will give an unannounced test during any class meeting is 1/5.If a student is absent twice,then the probability that the student will miss at least one test is
A.4/5
B.2/5
C.7/75
D.9/25
E.8/25
That is: P(Event A occurs) = 1 - P(Event A does not occur)
So, P(miss at least 1 test) = 1 - P(miss no tests)
P(miss no tests)
P(miss no tests) = P(no test on day 1 AND no test on day 2)
= P(no test on day 1) X P(no test on day 2)
= (4/5) X (4/5)
= 16/25
So, P(miss at least 1 test) = 1 - P(miss no tests)
= 1 - 16/25
= [spoiler]9/25 = D[/spoiler]
Cheers,
Brent