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ap183
08-04-2010, 12:32 PM
A question in a tutorial I had was:

A bank manager is presented with the details of ten accounts of which two are overdrawn. If he selects three of the ten accounts at random, find the probability that his sample contains the two overdrawn accounts

I have the answer but I don't understand them?

Thanks

08-04-2010, 01:49 PM
How many possible combinations are there of three drawn accounts?

How many possible combinations are there where two of them are overdrawn?

Figure that out and take the ratio.

ap183
08-04-2010, 02:07 PM
There are 1 possible combination for the first one and 3 for the second. 3C3 and 3C2. The ratio is 3:1.

08-04-2010, 04:34 PM
I think you misunderstood what I was asking.

How many ways can you draw three accounts from a selection of 10? The answer is: 10*9*8=720

Now, how many ways can you draw three accounts where two out of three are overdrawn?

Mean Joe
08-04-2010, 04:38 PM
ap183:

Think of splitting the accounts into overdrawn (n=2) and other (m=8).

How many possible combinations are there where two of them are overdrawn?
Answer: From the overdrawn accounts (n=2) you want to choose 2 of them; from the other accounts (m=8) you want to choose 1 of them.

ap183
08-04-2010, 04:58 PM
I think you misunderstood what I was asking.

How many ways can you draw three accounts from a selection of 10? The answer is: 10*9*8=720

Now, how many ways can you draw three accounts where two out of three are overdrawn?

Ah ok, so you mean 10P3 = 720. so 3P2 = 6 ways?

08-04-2010, 06:46 PM
Ah ok, so you mean 10P3 = 720. so 3P2 = 6 ways?

Good try, but no. Work it out manually. There are three ways you can have a combination where two out of three draws are overdrawn accounts. They are:

1,1,0
1,0,1
0,1,1

Where the 1's stand for overdrawn accounts. What's the probability of each of those situations happening (i.e. how many ways can each of those situations occur)?

For 1,1,0: You can have 16 different combinations. Either overdrawn1, overdrawn2, and one of the remaining accounts (which make up 8); or overdrawn2, overdrawn1, and one of the remaining accounts (making up the other 8).

ap183
08-05-2010, 07:57 AM
If I am correct in understanding how you got that figure then
for 1,0,1 - 16 combinations, overdrawn1, one of the other accounts, overdrawn2, or, overdrawn2, one of the other accounts, overdrawn 1

and same for the last one. So the probability is 16 for each of them?

08-05-2010, 11:08 AM
Yes. There are 16 ways in which you can have each of the drawn outcomes. This means that there are a total of 3 x 16 = 48 ways that you can draw three times and have two overdrawn accounts.

Now, take the ratio of (ways you can have overdrawn accounts) / (total combination of draws).

Also, be careful when you use the word probability. Probability is thought of as a value between 0 and 1, so it couldn't be 16.

ap183
08-05-2010, 05:29 PM
P(2 overdrawn, then one not) = 48/720 = 0.067
Ah that makes a bit more sense than the solution that was given:

P(2 overdrawn, then one not) = (2/10) * (1/9) * (8/8) = 1/45
There are three possible sequences with 2 overdrawn and one not, so the required probability is 3 * (1/45) = 1/15 = 0.067