eaakbari wrote:The below question is a very easy question, but I am having a conceptual error which I hope someone can shed light on.


There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?

Answer - [spoiler]1/28[/spoiler]

The method I chose to solve it was reverse probability approach, but my answer is wrong and not matching with my probability approach solution


Method 1 : Probability approach

Prob of BOTH Bob n rachel - 2/8 * 1/7 = 1/28




Method 2 : Reverse Probability approach

Prob of Bob n rachel in the committee - 1 - P(BOTH NOT IN COMMITTEE)

P can be under 3 possibilities


None are chosen - 6/8 * 5/7

Bob is chosen - 1/8 * 6/7

Rachel is chosen - 1/8 * 6/7



There for P = 1 - (6/8 * 5/7 + 1/8 * 6/7 + 1/8 * 6/7) = 14/56


Where is my thought process faultering ?

P(both Bob and Rachel) = 1 - P(not both Bob and Rachel)
= 1 - [P(B or R first AND not B or R second) + P(neither B nor R first AND anyone second)]
= 1 - [(2/8)(6/7) + (6/8)(7/7)]

eaakbari wrote:

But according what you have stated, the order seems to matter. But order should not matter here.

Assuming that in "[P(B or R first AND not B or R second)", we get B first & anyone second
and in
"P(neither B nor R first AND anyone second)]", we get anyone first and B second.

It makes (B , ANYBODY) in the committee.
which is the same case, we are accounting for twice !!

Where is my thought process going wrong?

Other way of putting this together

if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people.

1 - [(2/8)(6/7) + (6/8)(7/7)]