probability-unable to reconcile ans. between two approaches

[cutlass]

I just cooked up this problem and I get different answers when I try w/
and w/o combinations. Where am I going wrong?

Form a 4 member team w/ and 4 girls and 3 boys. What is the prob there are atleast 2 girls in the team ?

With combinations
---------------------

P(atleast 2 girls) = (4c2 + 4c3 + 4c4)/7c4 [ (2 girls + 3 girls + 4 girls) / total no. of ways ]
= 11/35

W/o combinations
---------------------

P(atleast 2 girls) = 1 - ( P (0 girls) + P ( 1 girl) )
P (0 girls) = 0
P (1 girl) = 4 * (3/7)(2/6)(1/5)(4/4) [ BBBG,BBGB,BGBB,GBBB ]
= 4/35
P(atleast 2 girls) = 1 - 4/35 = 31/35


Any good source for higher-level probability/combination questions?

[guest]

Hi -

your second approach gives you the correct answer of 31/35. Your problem in the first approach is that you forgot to count the different boys in the numerator. For example, how many ways can you form a team with three girls? The correct answer is 4c3*3c1 = 12. The forth spot on the team can be filled with any of the three boys. So to fix the calculation below, it becomes P(g>=2) = (4c2*3c2+4c3*3c1+4c4*3c0)/7c4 = (6*3+4*3+1*1)/35 = (18+12+1)/35 = 31/35.

Hope this was helpful.

I just cooked up this problem and I get different answers when I try w/
and w/o combinations. Where am I going wrong?

Form a 4 member team w/ and 4 girls and 3 boys. What is the prob there are atleast 2 girls in the team ?

With combinations
---------------------

P(atleast 2 girls) = (4c2 + 4c3 + 4c4)/7c4 [ (2 girls + 3 girls + 4 girls) / total no. of ways ]
= 11/35

W/o combinations
---------------------

P(atleast 2 girls) = 1 - ( P (0 girls) + P ( 1 girl) )
P (0 girls) = 0
P (1 girl) = 4 * (3/7)(2/6)(1/5)(4/4) [ BBBG,BBGB,BGBB,GBBB ]
= 4/35
P(atleast 2 girls) = 1 - 4/35 = 31/35


Any good source for higher-level probability/combination questions?

[cutlass]

Thanks. I was able to figure it out once I brushed up some probability material.

[StaceyKoprince]

Glad you guys got it worked out on your own! Keep up the good work :)

[Big Al]

Hi Guys,

Could somebody please clarify something for me? Why, when working out the probability of 1 girl above:

4 * [(3/7)(2/6)(1/5)(4/4)]

do we have to multiply the whole thing by 4 to get 4/35? If I had done this myself I would have left it at 1/35.

We know that the 4th person MUST be a girl and hence we multiply by 4/4. So why is the extra step necessary? Is it because it doesn't matter WHICH girl we pick - any one will do? I'm a bit confused....I would have thought that knowing that the fourth person is definitely a female would have been enough.

Many Thanks

[Guest]

Before we do anything we need to calculate how many total combinations there are: 4 spots and 7 people to choose from:

7!
-------
4! (3!)

= 35 Total Combinations

This is the most important part of the problem. By figuring this out and understanding it, you will be able to gauge your next computations based on this number. Try to understand and visualize that there are 35 possible combinations of 4 different girls and 3 different boys

The possible combinations are (1G 3B.....2G 2B.....3G 3B.....4G 0B)

The problem asks us to find out the number of combinations that contain 2 girls OR 3 girls OR 4 girls. We can either find out how many combinations contain 1 girl and 3 boys (1G 3 B) and subtract 35 from it---or---find out the number of combos for (2G 2B) + (3G 3B) + (4G 0B).

The easy way: 1 GIRL 3 BOYS

Girls: A,B,C,D BOYS: X,Y,Z

The combinations for 1 girl and three boys are (A, XYZ)....(B,XYZ)....(C,XYZ)....(D,XYZ) = 4

The technical way is: 4girls choose 1girl

4!
---- = 4
1!(3!)


***AND***

3boys choose 3boys

3!
------ = 1
3!

4x1 = 4

Out of the 35 possible combos above there are only 4 that contain three boys. 35-4 =31

31/35

[rfernandez]

Big Al, the 4 comes from the four different ways that one girl and three boys can be chosen:

GBBB
BGBB
BBGB
BBBG

Each of these has the same probability, namely (3/7)(2/6)(1/5)(4/4). You would then ADD all of the probabilities because this is an OR situation. A shortcut is simply to multiply by 4.

[Big Al]

Brilliant, makes sense now. Thanks a lot.

[RonPurewal]

Brilliant, makes sense now. Thanks a lot.

great.

keep in mind:
whenever you're doing a "multiply successive probabilities" approach, such as the one employed in that post, ORDER MATTERS.
this means that you can't just multiply together, say, gggb; you have to account for all possible orders in which those events can occur.