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06 Jul 2026, 09:03
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Given 8 qualified candidates.
To form a 3-person committee.
Total number of such committees possible = 8C3 = 56
Now, Invalid committees:
Number of ways to choose (A, B) together and any one of the remaining 6 candidates = 1 * 6 = 6
Also, the number of ways to choose (C, D) together and any one of the remaining 6 candidates = 1 * 6 = 6
Since the committee is of 3 members only, any overlapping committee containing (A, B) & (C, D) together is not possible.
Thus, overlapping committee = 0
Hence, the total number of invalid committees = 6 + 6 + 0 = 12
And, total number of valid committees = 56-12 = 44
Thus, the probability of forming a valid committee = 44/56 = 11/14
Hence, X = 11 & Y = 14
To form a 3-person committee.
Total number of such committees possible = 8C3 = 56
Now, Invalid committees:
Number of ways to choose (A, B) together and any one of the remaining 6 candidates = 1 * 6 = 6
Also, the number of ways to choose (C, D) together and any one of the remaining 6 candidates = 1 * 6 = 6
Since the committee is of 3 members only, any overlapping committee containing (A, B) & (C, D) together is not possible.
Thus, overlapping committee = 0
Hence, the total number of invalid committees = 6 + 6 + 0 = 12
And, total number of valid committees = 56-12 = 44
Thus, the probability of forming a valid committee = 44/56 = 11/14
Hence, X = 11 & Y = 14
06 Jul 2026, 09:09
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Bunuel
lets calculate total no of possible committees = 8C3 = \frac{8! }{ 3! 5!} = 56
since we cannot have incompatible candidates in the committee
lets find the no of committees with incompatible candidates
1. A, B - since 2 positions are already filled, the remaining one can be filled 6 ways - 6 committees
2. C, D - since 2 positions are already filled, the remaining one can be filled 6 ways - 6 committees
since both pairs cannot be on the same 3 person committee
total no of committees possible with incompatible candidates = 6+6 = 12
so no of committees possible = 56 - 12 = 44
so probability of valid committee = \frac{44}{56} = \frac{11}{14}
X = 11, Y = 14
06 Jul 2026, 09:15
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Total number os ways to calculate the selection 3 from 8 = 8C3=8*7*6/3*2*1=56
No. of ways in which AB or CD are in same team = 6+6=12 so valid teams=56-12=44
so X/Y=44/56 = 11/14
X=11 Y=14
No. of ways in which AB or CD are in same team = 6+6=12 so valid teams=56-12=44
so X/Y=44/56 = 11/14
X=11 Y=14
