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Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one
of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:
(a) 0.245 (b) 0.25
(c) 0.254 (d) 0.55?
of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:
(a) 0.245 (b) 0.25
(c) 0.254 (d) 0.55?
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Among the examinees in an examination 30%, 35% and 45% failed in Stati...
Problem: Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only.
Solution:
Given,
- The percentage of examinees who failed in Statistics is 30%
- The percentage of examinees who failed in Mathematics is 35%
- The percentage of examinees who failed in at least one of the subjects is 45%
Let A be the event that an examinee failed in Statistics,
B be the event that an examinee failed in Mathematics, and
C be the event that an examinee failed in at least one of the subjects.
We need to find the probability that an examinee failed in Mathematics only, i.e., failed in Mathematics but not in Statistics.
Using the formula of conditional probability, we have
P(B|A') = P(BnA')/P(A')
where A' denotes the complement of event A, i.e., the event that an examinee did not fail in Statistics.
We know that P(C) = 45% = 0.45 and P(A) = 30% = 0.3.
Using the formula of probability of union, we have
P(AuB) = P(A) + P(B) - P(AnB)
where AnB denotes the intersection of events A and B, i.e., the event that an examinee failed in both Statistics and Mathematics.
We know that P(C) = P(AuB) = 0.45 and P(AnB) = P(A) + P(B) - P(AuB) = 0.3 + 0.35 - 0.45 = 0.2.
Therefore, P(B') = P(BnA') + P(BnA) = P(A') - P(AnB) + P(B) - P(AnB) = 1 - 2P(AnB) - P(B) = 1 - 2(0.2) - 0.35 = 0.25.
Finally, we have
P(B|A') = P(BnA')/P(A') = (P(B) - P(AnB))/P(A') = (0.35 - 0.2)/0.7 = 0.214.
Hence, the probability that an examinee failed in Mathematics only is P(B|A') = 0.214, which is closest to option (a) 0.245.
Solution:
Given,
- The percentage of examinees who failed in Statistics is 30%
- The percentage of examinees who failed in Mathematics is 35%
- The percentage of examinees who failed in at least one of the subjects is 45%
Let A be the event that an examinee failed in Statistics,
B be the event that an examinee failed in Mathematics, and
C be the event that an examinee failed in at least one of the subjects.
We need to find the probability that an examinee failed in Mathematics only, i.e., failed in Mathematics but not in Statistics.
Using the formula of conditional probability, we have
P(B|A') = P(BnA')/P(A')
where A' denotes the complement of event A, i.e., the event that an examinee did not fail in Statistics.
We know that P(C) = 45% = 0.45 and P(A) = 30% = 0.3.
Using the formula of probability of union, we have
P(AuB) = P(A) + P(B) - P(AnB)
where AnB denotes the intersection of events A and B, i.e., the event that an examinee failed in both Statistics and Mathematics.
We know that P(C) = P(AuB) = 0.45 and P(AnB) = P(A) + P(B) - P(AuB) = 0.3 + 0.35 - 0.45 = 0.2.
Therefore, P(B') = P(BnA') + P(BnA) = P(A') - P(AnB) + P(B) - P(AnB) = 1 - 2P(AnB) - P(B) = 1 - 2(0.2) - 0.35 = 0.25.
Finally, we have
P(B|A') = P(BnA')/P(A') = (P(B) - P(AnB))/P(A') = (0.35 - 0.2)/0.7 = 0.214.
Hence, the probability that an examinee failed in Mathematics only is P(B|A') = 0.214, which is closest to option (a) 0.245.
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Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55?
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Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55?.
Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only:(a) 0.245 (b) 0.25(c) 0.254 (d) 0.55?.
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