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26 Apr 2026, 14:03
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In a class of 48 students, some are in chess club and some are in swim class. If one student is selected at random, what is the probability that the student is in chess club?
(1) The probability that the student selected is in chess club or in swim class is \(\frac{11}{12}\).
(2) The probability that the student selected is in swim class is \(\frac{5}{6}\).
(1) The probability that the student selected is in chess club or in swim class is \(\frac{11}{12}\).
(2) The probability that the student selected is in swim class is \(\frac{5}{6}\).
26 Apr 2026, 14:03
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Official Solution:
In a class of 48 students, some are in chess club and some are in swim class. If one student is selected at random, what is the probability that the student is in chess club?
Total = Chess + Swim - Both + Neither
\(48 = Chess + Swim - Both + Neither\)
We want to find \(\frac{Chess}{48}\).
(1) The probability that the student selected is in chess club or in swim class is \(\frac{11}{12}\).
This implies Chess + Swim - Both = \(\frac{11}{12} * 48 = 44\). But this does not determine how many are in chess club. Not sufficient.
(2) The probability that the student selected is in swim class is \(\frac{5}{6}\).
This implies Swim = \(\frac{5}{6} * 48 = 40\). But this does not determine how many are in chess club. Not sufficient.
(1)+(2) We have Chess + Swim - Both = 44 and Swim = 40. Thus, Chess + 40 - Both = 44, which gives Chess - Both = 4. Since we still do not know how many are in both swim class and chess club, we cannot determine how many are in chess club. Not sufficient.
Answer: E
In a class of 48 students, some are in chess club and some are in swim class. If one student is selected at random, what is the probability that the student is in chess club?
Total = Chess + Swim - Both + Neither
\(48 = Chess + Swim - Both + Neither\)
We want to find \(\frac{Chess}{48}\).
(1) The probability that the student selected is in chess club or in swim class is \(\frac{11}{12}\).
This implies Chess + Swim - Both = \(\frac{11}{12} * 48 = 44\). But this does not determine how many are in chess club. Not sufficient.
(2) The probability that the student selected is in swim class is \(\frac{5}{6}\).
This implies Swim = \(\frac{5}{6} * 48 = 40\). But this does not determine how many are in chess club. Not sufficient.
(1)+(2) We have Chess + Swim - Both = 44 and Swim = 40. Thus, Chess + 40 - Both = 44, which gives Chess - Both = 4. Since we still do not know how many are in both swim class and chess club, we cannot determine how many are in chess club. Not sufficient.
Answer: E
07 May 2026, 01:09
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we can use below relation to solve this question:
P(A or B)=P(A) + P(B) - P(A and B)
Now consider P(A)= probability that the student is in chess club
P(B)= probability that the student is in swim club
P(A or B)= probability that the student is in chess club or in swim club
P(A and B)= probability that the student is in chess club and swim club (student is part of both club)
and we need to find P(A). To find P(A) we will need P(A or B), P(B) and P(A and B).
(i) as per above relation, statement 1 only provide info. about student who are part of chess club or swim club i.e. P(A or B) so not sufficient.
(ii)this statement only provide info. about student who are part of swim club i.e. P(B) so not sufficient.
combining i and ii , we still not have information about P(A and B) so answer is E
I'll also suggest to go through Bunuel's explanation to know how to deal with numbers in such scenario.
P(A or B)=P(A) + P(B) - P(A and B)
Now consider P(A)= probability that the student is in chess club
P(B)= probability that the student is in swim club
P(A or B)= probability that the student is in chess club or in swim club
P(A and B)= probability that the student is in chess club and swim club (student is part of both club)
and we need to find P(A). To find P(A) we will need P(A or B), P(B) and P(A and B).
(i) as per above relation, statement 1 only provide info. about student who are part of chess club or swim club i.e. P(A or B) so not sufficient.
(ii)this statement only provide info. about student who are part of swim club i.e. P(B) so not sufficient.
combining i and ii , we still not have information about P(A and B) so answer is E
I'll also suggest to go through Bunuel's explanation to know how to deal with numbers in such scenario.
