MRLouis wrote:

Hello everyone! Can someone please help to solve the question below?

Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacemet from thi group, what is the probability tht Mark and not Anna will be selected?

- 1/7

- 2/7

- 3/7

- 4/7

- 5/7

Thank you very mcuh!

Louis

Dear

MRLouis,

I'm happy to respond.

This problem can be addressed relatively easily by using probability counting techniques. See:

GMAT Probability and Counting TechniquesQuestion #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

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)