Anna and Mark are included in a group of 8 people. If 4 people are

Question

Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the 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

mikemcgarry · Accepted Answer

Dear MRLouis , I'm happy to respond.

This problem can be addressed relatively easily by using probability counting techniques. See: GMAT Probability and Counting Techniques Question #1 : What's the total number of groups that could be drawn here? That's given by: 8C4 = \dfrac{8!}{(4!)(4!)} = \dfrac{8*7*6*5}{(4*3*2*1)} = 2*7*1*5 = 70 That's the denominator. Question #2 : How many groups would satisfy the condition? We would have Mark and then could choose 3 from the remaining 6, not including Anna. That's given by: 6C3 = \dfrac{6!}{(3!)(3!)} = \dfrac{6*5*4}{(3*2*1)} = 5*4 = 20 That's the numerator of the probability fraction. Probability = \dfrac{20}{70} = \dfrac{2}{7} Answer = (B) Does all this make sense? Mike

Bunuel · Answer

We need {Mark, Any but Anna, Any but Anna, Any but Anna} in any order.

P(MAAA) = \frac{4!}{3!}*\frac{1}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}=\frac{2}{7} . We multiply by 4!/3! = 4 because {Mark, Any but Anna, Any but Anna, Any but Anna}, MAAA, can be arranged in four ways: MAAA, AMAA, AAMA, AAAM.

Answer: B.

Bunuel · Answer

Who can find logic behind the following solution:

\frac{1 - (\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*1*1 + \frac{6}{8}*\frac{5}{7}*\frac{4}{7}*\frac{3}{5})}{2}=\frac{2}{7}

hoang221 · Answer

The numerator of the left equation can be understand as: 1 - ( probability that both are selected + probability both are not selected) and it's equal to (only 1 of them is selected), we need to divide it by 2 because the numerator is account for both situation, where only Anna is chosen or only Mark is chosen, and we only want Mark.

JeffTargetTestPrep · Answer

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!/  = (6 x 5 x 4)/(3 x 2 x 1) = 20.

The number of ways to select 4 people from 8 is 8C4 = 8!/  = (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

Crater · Answer

There are 8 people: Mark, Anna, Person 3, Person 4, Person 5, Person 6, Person 7, Person 8 
 1. First select Mark: 1/8
2. Now not select Anna: (1-1/7) = 6/7
3. Now select any except Anna: 5/6
4. Now select any except Anna: 4/5

Now 1, 2, 3, 4 can be arranged in 4!/3! ways because 2, 3, 4 are identical

Therefore, required Prob:
 \frac{4!}{3!}*\frac{1}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{2}{7}

Hence, Option (B)

Rakshit25 · Answer

Another way of solving this:

Let's say Mark is already included in the group, then the number of ways of choosing the remaining 3 people such that Anna is not a part of the group is 6C3=20

The total number of ways of selecting 4 people from a group of 8 is 8C4=70

Thus, the probability that Mark is selected and Anna is not = 20/70=2/7

PSKhore · Answer

We need to select Mark always, therefore probability of selecting Mark = 1
The other 3 people excluding Anna and Mark can be selected in 6C3 ways = 20
Total ways of selecting 4 people from 8 people = 8C4 = 70

Therefore total ways = 20/70
=2/7

Anna and Mark are included in a group of 8 people. If 4 people are

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Anna and Mark are included in a group of 8 people. If 4 people are

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615 to 715 in 15 Days (Ria's Strategies)

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