# Probability Involving Combinations and Permutations

• Mar 6th 2012, 11:50 AM
GrigOrig99
Probability Involving Combinations and Permutations
I think I have worked this out correctly but wanted to have someone else take a look to confirm it.
Q.: A box contains letters used in a word game. At a certain stage in the game, the 8 letters in the box are A, A, C, E, L, P, P, P. 1 player draws, at random, 3 letters in succession without replacing them. Calculate the probability that the 3 letters drawn do not include E or P.

Attempt: P(3 letters without E or P) = (2/8 x 1/7 x 1/6) x 3! = 12/336 = 1/28, P(3 letters without P or E) = (2/8 x 1/7 x 1/6) x 3! = 1/28. Thus, P(3 letters without E/ P) = 1/28 + 1/28 = 1/14.

This matches the text book answer. I have separated the probabilities as 'P or E' and 'E or P' due to appearance of the 'or' in the question which, according to the book, means the answer totals are to be added.

Many thanks.
• Mar 6th 2012, 12:16 PM
Plato
Re: Probability Involving Combinations and Permutations
Quote:

Originally Posted by GrigOrig99
I think I have worked this out correctly but wanted to have someone else take a look to confirm it.
Q.: A box contains letters used in a word game. At a certain stage in the game, the 8 letters in the box are A, A, C, E, L, P, P, P. 1 player draws, at random, 3 letters in succession without replacing them. Calculate the probability that the 3 letters drawn do not include E or P.
Attempt: P(3 letters without E or P) = (2/8 x 1/7 x 1/6) x 3! = 12/336 = 1/28, P(3 letters without P or E) = (2/8 x 1/7 x 1/6) x 3! = 1/28. Thus, P(3 letters without E/ P) = 1/28 + 1/28 = 1/14.

Are you saying that the book gives the answer as $\displaystyle \frac{1}{14}~?$

Think about it. The only draws that do not contain a P or E is
$\displaystyle \{A,C,L\},~\{A,A,C\},~\{A,A,L\}$
What is the probability of each of those?
• Mar 6th 2012, 12:40 PM
GrigOrig99
Re: Probability Involving Combinations and Permutations
Ok, so {A, C, L}, which can be arranged 3! = (2/8 x 1/7 x 1/6) x 3! = 2/336 x 3! = 12/336, {A, A, C}, which can be arranged 3 ways = (2/8 x 1/7 x 1/6) x 3 = 6/336 and {A, A, L}, which can arranged 3 ways = (2/8 x 1/7 x 1/6) x 3 = 6/336.
Therefore, 12/336 + 6/336 + 6/336 = 24/336 = 1/14.
Thanks again.
• Mar 6th 2012, 01:33 PM
Plato
Re: Probability Involving Combinations and Permutations
Quote:

Originally Posted by GrigOrig99
Ok, so {A, C, L}, which can be arranged 3! = (2/8 x 1/7 x 1/6) x 3! = 2/336 x 3! = 12/336, {A, A, C}, which can be arranged 3 ways = (2/8 x 1/7 x 1/6) x 3 = 6/336 and {A, A, L}, which can arranged 3 ways = (2/8 x 1/7 x 1/6) x 3 = 6/336.
Therefore, 12/336 + 6/336 + 6/336 = 24/336 = 1/14.
Thanks again.

You may want to think about why this works.
$\displaystyle \dfrac{\binom{4}{3}}{\binom{8}{3}}=\dfrac{1}{14}$
• Mar 6th 2012, 02:05 PM
GrigOrig99
Re: Probability Involving Combinations and Permutations
The 4 in (4 nCr 3) refers to the 4 viable letters A, A, C, L and the 3 is for all possible combinations using just 3 of these letters. Similarly, the 8 from (8 nCr 3) refers to all the letters, including E and P, and the 3 is for all possible combinations using just 3 of all these letters. This will act as the total from which we derive the fraction of successful events, i.e. (4 nCr 3).