Post a New Question

Algebra

posted by on .

From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).

It might be a little out of order but this is what I got.

C(10,2)
(10*9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1*2*1)
=45

C(6,2)
(6*5*4*3*2*1)/(2*1)
=360

C(4,2)
(4*3*2*1)/(2*1
=12

360*12 = 4320

s/s+f
4320/4320+45
4320/4365

99% answer

  • Algebra - ,

    no how about this

    P(at least 2 women)=
    P(exactly 2 women)+P(exactly 3 women)=
    {(6!/(1!*5!))*(4!/(2!*2!))+
    (4!/(3!*1!)}/{10!/(3!*7!)}=
    40/120=1/3

    answer = 1/3

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question