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In a group of 80 students, 24 are enrolled in geometry, 40 in biology
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23 Aug 2021, 04:28
23 Aug 2021, 04:28
Expert Reply
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Question Stats:
83% (01:11) correct
16% (01:21) wrong based on 12 sessions
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In a group of 80 students, 24 are enrolled in geometry, 40 in biology, and 20 in both. If a student were randomly selected from the 80 students, what is the probability that the student selected would not be enrolled in either course?
(A) 0.20
(B) 0.25
(C) 0.45
(D) 0.55
(E) 0.60
(A) 0.20
(B) 0.25
(C) 0.45
(D) 0.55
(E) 0.60
In a group of 80 students, 24 are enrolled in geometry, 40 in biology
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24 Aug 2021, 03:44
24 Aug 2021, 03:44
1
Expert Reply
The probability of occurrence of two events A or B, happening given that they are not mutually exclusive i.e \( P (A \cap B) \neq 0 \) is given as:
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = P(A) + P(B) - P(A and B)
Also the probability of an event E not occurring is P(E') = 1 - P(E)
Lets provide a breakdown of the events here:
Total Students = 80
P(Geo) = \(\frac{ 24}{80 }\)
P(Bio) = \(\frac{ 40}{80 }\)
P(Both) = \(\frac{ 20}{80 }\)
P(Geo or Bio) = P(Geo) + P(Bio) - P(Both)
\(\implies\) P(Geo or Bio) = \(\frac{ 24}{80 } + \frac{ 40}{80 } - \frac{ 20}{80 }\)
\(\implies\) P(Geo or Bio) = \(\frac{ 24 + 40 - 20}{80} = \frac{ 44}{80} = \frac{ 11 }{ 20 } = 0.55 \)
The question requires us to pick a student who has not enrolled in either of these courses:
\(\implies \) P(Both') = 1 - P(Both) = 1 - 0.55 = 0.45
Thus the answer is Option C
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Re: In a group of 80 students, 24 are enrolled in geometry, 40 in biology
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24 Aug 2021, 09:14
24 Aug 2021, 09:14
1
Carcass wrote:
In a group of 80 students, 24 are enrolled in geometry, 40 in biology, and 20 in both. If a student were randomly selected from the 80 students, what is the probability that the student selected would not be enrolled in either course?
(A) 0.20
(B) 0.25
(C) 0.45
(D) 0.55
(E) 0.60
(A) 0.20
(B) 0.25
(C) 0.45
(D) 0.55
(E) 0.60
Total = Geometry + Biology - Both + Leftovers
80 = 24 + 40 - 20 + Leftovers
Leftovers = 36
Probabilty = \(\frac{36}{80} = 0.45\)
Hence, option C
