Problem Solving

Can someone please explain how to solve the below. Many thanks!

From a group of 8 volunteers including Andrew and Karen, 4 volunteers are to be selected at random. What is the probability that Andrew will be among the 4 selected and Karen will not?

From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?

A 3/7
B 5/12
C 27/70
D 2/7
E 9/35

Alternate approach:

From the 8 people, 4 will be selected.
Thus, P(A is selected) = 4/8.
From the 7 remaining people, 4 will NOT be selected.
Thus, P(K is not selected) = 4/7.
To combine these probabilities, we multiply:
4/8 * 4/7 = 2/7.

The correct answer is D.

From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?

A 3/7
B 5/12
C 27/70
D 2/7
E 9/35

Here's even another approach.

First the probability that Andrew is selected is 1 - the probability that Andrew is not selected.

The probability that Andrew is not selected can work like this.

The first person chosen has 7/8 chance of being not Andrew, the second has 6/7, the third, 5/6, and the fourth 4/5. If that's how the four are selected, then no Andrew.

Multiply them out to get (7/8)(6/7)(5/6)(4/5) = 4/8 = 1/2. So half the time Andrew is not selected, and 1 - 1/2 = 1/2 of the time Andrew is selected.

So we know the chances of Andrew being selected, 1/2.

Now what is the chance that with Andrew selected Karen is not.

With Andrew already selected, we have three more slots to fill, with seven people left to fill them. The first has a 6/7 chance of being not Karen, the second, 5/6, and the third, 4/5. Multiply them to get (6/7)(5/6)(4/5) = 4/7 chance of Karen not being selected from the other 7.

So we have the chance that Andrew will be selected, 1/2, and the chance that from the rest Karen will not be selected, 4/7. So to get the probability that Andrew will be selected and Karen will not be selected, multiply those to get (1/2)(4/7) = 2/7.

Choose D.

P.S. It's kinda bugging me that with Andrew selected there is not a 1/2 chance that Karen is not selected. I mean originally I figured that with Andrew selected, Karen would be selected half the time and not selected the other half of the time. So this would be an easy problem. But that's not the case, and I was saved by their not having put a trap answer that would have worked that way.

P.P.S. Gotta love the way Brent did it because it is so basic and is applicable to so many situations. No need to have all kinds of different strategies at hand. Just figure out how many ways one can get the favorable outcome and divide it by the total number of possible outcomes. End of story.

So cool to see so many cool methods and ways of thinking here.

I approach this problem by first working how many combinations of 4 people there are: this is 8C4, which is 8!/(4!.4!) or 70.
Now I need to work out how many combinations there are with Adam in and Karen out. We can set Andrew in and then see how many combinations of 3 people there without being able to chose Karen. This is 6C3, 6!/(3!.3!) or 20.

Thus, the answer is 20/70 or 2/7.

Brent's way is safe - here's another quick way.

Let's start with the probability that Andrew is selected: 1/2. (Half the people are selected, half are not.)

Now that Andrew is in, we need to worry about Karen. Since we know Andrew is in, 4 of the 7 remaining people will be out. The probability that Karen is one of them is thus 4 out of 7.

Since we need BOTH of these to occur, our probability = 1/2 * (4/7), or 2/7.