Statistics question please help?

Question

The question is "what is the probability of selecting 2 males and 5 females from a committee of 6 males and 11 females." I don't even know where to begin to figure out this problem.

Answer 1

total number of persons= number of males+ number of females
= 6+11= 17
total number ogf persons to be selected= 2+5=7
7 persons out of 17 can be selected in=17C7 ways
=17*16*15*14*13*12*11 divided by 7*6*5*4*3*2*1
= 19448 ways { after solving}
2 males out of 6 males can be selected in= 6C2 ways
= 6*5 divided by 2*1
= 15 ways
5 females out of 11 can be selected in= 11C5 ways
= 11*10*9*8*7 divided by 5*4*3*2*1
= 462 ways

so probability of selecting 2 males and 5 females from a committee of 6 males and 11 females = 462*15 divided by 19448
= 6930/19448
= 3465/9724

Answer 2

In the experiment of selecting a males and females from the commitee.

Total no possible choices = No of ways in which 7 members can be selected from total of 17

==> n = 17C7
==> n = (17x16x15x14x13x12x11) / (7x6x5x4x3x2x1)
==> n = (98017920) / (5040)
==> n = 19448

Let A be the event of selecting 2 males and 5 females from 6 men and 11 females respectively.

For event A
----------------

males females total
-------- ----------- --------
available 6 11 17
selct 2 5 7
-------- ----------- --------
6C2 11C5 17C7

No of favourable choices for selecting 2 males and 5 females from 6 males and 11 females respectively
= 6C2 x 11C5
= [ (6x5) / 2 ] x [(11x10x9x8x7x6) / (5x4x3x2x1)]
= [ (30) / 2 ] x [(55440) / (120)]
= [ 15 ] x [ 462 ]
= [ 15 x 462 ]
m = 6930

The probability of occurance of an event A = No of favourable choices / Total no of possible choices

==> P(A) = m / n
==> P(A) = 6930 / 19448
==> P(A) = 0.35633484162895927601809954751131


For more detailes and clear understanding:

http://www.futureaccountant.com/probability/

http://www.schoolingkids.com/probability/