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07 Jul 2026, 04:51
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Here, total pool of qualified candidate = 8
Company need to form a three-person leadership committee and committee will be valid if its three members do not include either of these pairs of incompatible candidates.
Total Possible Ways 8C3 = (8 x 7 x 6 x 5!) / ( 5! x 3 * 2 * 1) = 56
Now a committee of three member needs to be formed and
As we know that Aly and Ben, and Cam and Dan have conflict issue.
and we need to find the probability that the selected candidates form a valid committee is X / Y
So in how many ways a invalidate committees can be formed
Assume Aly and Ben both are in a committee, in this way total 6 committees can be formed
Similarly with Cam and Dan total 6 invalid committees can be formed. So total 12 invalid committees
So total valid committees are 56 - 12 = 44
The probability that the selected candidate form a valid committee = 44 / 56 = 11 / 14 (simplified form)
Company need to form a three-person leadership committee and committee will be valid if its three members do not include either of these pairs of incompatible candidates.
Total Possible Ways 8C3 = (8 x 7 x 6 x 5!) / ( 5! x 3 * 2 * 1) = 56
Now a committee of three member needs to be formed and
As we know that Aly and Ben, and Cam and Dan have conflict issue.
and we need to find the probability that the selected candidates form a valid committee is X / Y
So in how many ways a invalidate committees can be formed
Assume Aly and Ben both are in a committee, in this way total 6 committees can be formed
Similarly with Cam and Dan total 6 invalid committees can be formed. So total 12 invalid committees
So total valid committees are 56 - 12 = 44
The probability that the selected candidate form a valid committee = 44 / 56 = 11 / 14 (simplified form)
07 Jul 2026, 05:31
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To find the probability, we need to divide Suitable arrangements / Total arrangements.
\(Total = C^3_8 = \frac{8! }{ 3!5!} = 56 =\) all combinations of 3 out of 8.
Now for Suitable: I suggest simply subtracting all the ones that don't fit from the total.
All committees with AB together are: ABC - ABD - ABE - ABF - ABG - ABH, which makes 6 total.
Same number goes for CD together, which is 6 more.
So, all Favourable committees are \(Total - 6 - 6 = 56 - 12 = 44\)
Finally, the probability is \(\frac{44}{56} = \frac{11}{14}\), so the answer is \(x=11\), and \(y=14.\)
\(Total = C^3_8 = \frac{8! }{ 3!5!} = 56 =\) all combinations of 3 out of 8.
Now for Suitable: I suggest simply subtracting all the ones that don't fit from the total.
All committees with AB together are: ABC - ABD - ABE - ABF - ABG - ABH, which makes 6 total.
Same number goes for CD together, which is 6 more.
So, all Favourable committees are \(Total - 6 - 6 = 56 - 12 = 44\)
Finally, the probability is \(\frac{44}{56} = \frac{11}{14}\), so the answer is \(x=11\), and \(y=14.\)
07 Jul 2026, 05:35
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Total possible committees of three people out of 8: 8c3 -> 56
Invalid committees: if both AB or both CD in the committee together->
So, scenarios:
1. In the first two positions, AB - fixed;
Last position -> 6 people can come.(6c1)
2. Similarly -> In the first two positions, CD - fixed;
Last position -> 6 people can come.(6c1)
So valid committee: (Total possible committee - invalid committee) = 56-12 = 44
probability: 44/56 = 11/14
x = 11, y = 14
Invalid committees: if both AB or both CD in the committee together->
So, scenarios:
1. In the first two positions, AB - fixed;
Last position -> 6 people can come.(6c1)
2. Similarly -> In the first two positions, CD - fixed;
Last position -> 6 people can come.(6c1)
So valid committee: (Total possible committee - invalid committee) = 56-12 = 44
probability: 44/56 = 11/14
x = 11, y = 14
