# Combinatorial Probability

• January 22nd 2011, 10:52 AM
NOX Andrew
Combinatorial Probability
I'm experiencing difficulty solving one of the exercises in my probability and statistics. Any help would be appreciated.

Exercise 1-29 At a picnic, there was a bowl of chocolate candy that had 10 pieces each of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.

(b) What is the probability that Jen grabs exactly five varieties?

I know there are ${60\choose6}$ combinations of grabbing 6 pieces of candy at random from the bowl of 60 chocolate candies. I believe there are ${10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {45\choose1}$ outcomes in which Jen grabs exactly five varieties. Therefore, the probability that Jen grabs exactly five varieties is approximately 0.089885. However, the solutions manual gives the answer as 0.05393.
• January 22nd 2011, 11:04 AM
Quote:

Originally Posted by NOX Andrew
I'm experiencing difficulty solving one of the exercises in my probability and statistics. Any help would be appreciated.

Exercise 1-29 At a picnic, there was a bowl of chocolate candy that had 10 pieces each of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.

(b) What is the probability that Jen grabs exactly five varieties?

I know there are ${60\choose6}$ combinations of grabbing 6 pieces of candy at random from the bowl of 60 chocolate candies. I believe there are ${10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {45\choose1}$ outcomes in which Jen grabs exactly five varieties. Therefore, the probability that Jen grabs exactly five varieties is approximately 0.089885. However, the solutions manual gives the answer as 0.05393.

But there are $\binom{6}{5}$ ways to choose 5 of the 6 varieties,
• January 22nd 2011, 11:12 AM
Failure
Quote:

Originally Posted by NOX Andrew
I'm experiencing difficulty solving one of the exercises in my probability and statistics. Any help would be appreciated.

Exercise 1-29 At a picnic, there was a bowl of chocolate candy that had 10 pieces each of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.

(b) What is the probability that Jen grabs exactly five varieties?

I know there are ${60\choose6}$ combinations of grabbing 6 pieces of candy at random from the bowl of 60 chocolate candies. I believe there are ${10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {10\choose1} \times {45\choose1}$ outcomes in which Jen grabs exactly five varieties. Therefore, the probability that Jen grabs exactly five varieties is approximately 0.089885. However, the solutions manual gives the answer as 0.05393.

First, count the number of ways to choose the 5 varieties that he picks only once, then multiply this by the number of ways that he can choose a single element from those 5 varieties, then multiply this by the number of ways to choose two elements from the single remaning variety. This gives the numerator of the following:

$\displaystyle\frac{\binom{6}{5}\cdot \binom{10}{1}^5\cdot \binom{10}{2}}{\binom{60}{6}}\approx 0.05393$
• January 22nd 2011, 11:39 AM
NOX Andrew
It seems odd to count the number of ways to choose the 5 varieties. I thought the purpose of combinations was to eliminate the different order in which the candies are chosen. But I can live with that. I'm entirely lost as to why one would count the number of ways to choose two elements from the single remaining variety. It seems we shouldn't consider the remaining variety since Jen is supposed to have only five varieties, not six.
• January 22nd 2011, 11:45 AM
Failure
Quote:

Originally Posted by NOX Andrew
I'm entirely lost as to why one would count the number of ways to choose two elements from the single remaining variety. It seems we shouldn't consider the remaining variety since Jen is supposed to have only five varieties, not six.

Well, Jen did choose 6 pieces in total, since he has got pieces from exactly 5 different varieties he must have chosen 2 pieces from the same (remaining) variety.
• January 22nd 2011, 11:52 AM
Failure
Quote:

Originally Posted by NOX Andrew
It seems odd to count the number of ways to choose the 5 varieties. I thought the purpose of combinations was to eliminate the different order in which the candies are chosen. But I can live with that.

Sorry, I forgot to answer this objection in my first reply. Here is why: the way I am counting is not meant to count the different orders in which Jen chooses his 6 pieces (one after another), it is only meant to count the number of different ways the final combination of 5 pieces from 5 different varieties and 2 pieces from the single remaining variety can be done - this is quite independent of the particular order in which the 6 pieces have been chosen.
• January 22nd 2011, 11:52 AM
NOX Andrew
That makes sense. I misunderstood the question in that regard. Thank you for help.
• January 22nd 2011, 12:09 PM
Plato
There is a minor typo.
It should be $\displaystyle\frac{\binom{6}{5}\cdot \binom{10}{1}^4\cdot \binom{10}{2}}{\binom{60}{6}}\approx 0.05393$.
• January 22nd 2011, 01:39 PM
Is it not...

1. Pick 4 varieties that Jen chooses once

2. Count the number of ways to have 1 from each

3. Choose 1 of the remaining two varieties from which to choose the final 2.

This gives a numerator of $\displaystyle\binom{6}{4}\;10^4\;\binom{2}{1}\;\bi nom{10}{2}$

For example, if we label the varieties A, B, C, D, E, F

then the $\binom{6}{5}$ ways to choose 5 varieties are

ABCDE, ABCDF, ABCEF, ABDEF, ACDEF, BCDEF.

Taking ABCDE...

the pair can come from any of these 5 and the 4 distinct ones from the other 4.

This gives a probability numerator of $\displaystyle\binom{5}{1}\binom{10}{1}^4\;\binom{1 0}{2}\binom{6}{5}$

Is the book answer correct ?
• January 22nd 2011, 02:22 PM
Plato
Actually I agree with you, I think the book's answer is probably wrong.
I get the same as you by a different means: six way to pick flavors, five ways to pick the one that doubles, then $10^4\cdot \binom{10}{2}$.
I simply pointed out that to get the given answer the exponent had to be 4 not 5.