aNNA and mark

if out of 4 people selected Mark is already selected (probability: 1) we need to select the rest 3 people from the remaining 6 people (8-mark-anna)


so 1*6C3/8C4=2/7

hence, B

can someone explain this exercise with another way. I'm interesting in know how to do with conditional probability.

figs wrote:can someone explain this exercise with another way. I'm interesting in know how to do with conditional probability.

Let E be the event that the first 3 persons selected are neither Mark nor Ann

then M be the event of selecting M

A be the event of selecting A. Then A^c be the event she is not selected = 1-A

Probability of selecting 4 people = Number of ways of selecting first 3 x number of ways of selecting Mark given that Anna is not selected.
.
P(E)= 6C3/8/C4

P(M|A^c) = means that of the remaining 5 persons to choose from, Anna cannot be chosen and so cannot the others b/c we must have Mark.

Let's evaluate P(M|A^c) which you request.
P(M|A^c) = P(M intercept A^c) /P(A^c)

Given the indepdence of the events, this becomes
P(M|A^c) = P(M intercept A^c) /P(A^c) = P(M) P(A^c)/P(A^c) =P(M)

Now we know P(M) is a certainty since we already have the other three people and the probabability of certainty is 1.

That was the solution he earlier provided. I am just giving it the explanation you wanted.

Simpler solution:

Probability mark will be selected: 1/2

Probability Anna WILL be selected = 6/7*5/6*4/5*3/4 = 3/7 (no need to actually do the calculation: 3 people out of 7 will be selected, so it's just directly 3/7)


Probability Anna WILL NOT be selected is therefore 1 - 3/7 = 4/7

Multiply the two: 1/2 * 4/7 = 2/7

There's no reason to dabble in combinatorics for this kind of stuff.

figs wrote:Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from this group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7

The calculation of the number of ways to select Mark but not Anna requires that we only consider the 6 remaining individuals because we don’t count Anna at all, and we already know that Mark has been selected. Thus, there are 6 available individuals to fill just 3 slots: 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/(3 x 2 x 1) = 20.

The number of ways to select 4 people from 8 is 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 7 x 2 x 5 = 70.

So the probability that Mark will be selected and Anna will not be selected is 20/70 = 2/7.

Answer: B