Trying to understand probabilities better...

mayonnai5e: Legendary Member; Posts: 1018; Joined: Tue Dec 12, 2006 7:19 pm; Thanked: 86 times; Followed by:6 members

by mayonnai5e » Wed Sep 12, 2007 3:24 pm

I can solve basic probability questions and even some more advanced ones, but the problem is it takes too long because I often start off on the wrong track and get confused by how to set up the probability formula. Sometimes I try to multiply several probabilities and other times I try to use the standard formula, but use the wrong calculations. I have a few questions that I hope some of the probability gurus on this forum can help me with and some examples to look at (examples were posted in other threads):

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

If the ans is D ie 3/5 then my explanation goes like this:-
Total no of events = Selecting 4 children from a group of 6
= 6C4 = 15
Favorable events will only be when 2 boys and two girls are selected (Since equal nos)
Selecting 2 Girls from a group of 3 Girls = 3C2 = 3
Selecting 2 Boys from a group of 4 Boys = 3C2 = 3
Total favorable events = 3x3 = 9
Prob = 9/15
or 3/5

In this case, the solution involved using the std prob formula with the combination formula.

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with the correct address?

a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

Probability that only 1 letter will be put into the envelope with the correct address is (d) 1/3

Step 1 - No. of ways you can select one letter (out of the four given) = 4
and there is only one way to put this letter into the correct envelope.

Now we need to ensure that none of the other letters (B,C and D) are put into the correct envelopes.

Step 2 - So, the next letter, can be put in an incorrect envelope in 2! ways (because one envelope out of the remaining three envelopes will be the correct one for this letter and hence we cannot use that one)

Step 3 - Now there are two letters remaining and there is only one way to put them into incorrect envelopes.

Required Probability = (4 * 2)/ 4! [Total number of ways is 4!]
= 1/3

In this second case, the std formula is used with permutations.

So my first question is, how can I easily determine whether to use combinations or permutations?

My next question is how not to get off track. I sometimes make this mistake (another forum member posted this in response to the first question:

Ok may be I am just getting very confused but here is what I did and I got the ans as 1/10

To get equal no of boys and girls u need to find the probability of selecting 2 girls and the probability of selecting 2 boys.

Probability of selecting a girl on the first chance is 3/6

and the prob of selecting a girl in the 2nd chance is 2/5

and the prob of selecting a boy in the 3rd round is 3/4

and the prob if selecting a boy in the 4th round is 2/3

So the prob of getting equal girls and boys is 3/6*2/5*3/4*2/3 = 1/10

Can some one pls tell me where I am going wrong???

I make the same mistake sometimes and first try to multiply the individual probabilities and then decide afterwards that something isn't right when I notice the answers choices do not contain my answer, then I go back and rethink the whole problem. This takes a long time!

Thanks in advance.

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