Probability of Combinations and Permutations

[mdzvi]

Hi, I'm a mom trying to help my son understand why he got answers wrong on his online math program.
He is taking Geometry, but the last unit in the class is an introduction to Probability and Statistics.

After re-reviewing the lesson and re-working the problems he got wrong, we were able to figure out most of the correct answers (however, it seems as if the big glitch is understanding when the sample space is a Combination vs a Permutation). However, we can't figure out 1 question no matter what methodology we try.

If someone could help explain the problem below, we would really appreciate it.
Also, if anyone has a suggestion for additional online learning modules that cover this topic, that would be great.
I've tried looking, but am not finding anything that explains it in a similar way to the online lesson. Thanks.

Here is the problem:

A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district debate. What is the probability that 1 girl and 2 boys will be selected?

The methodology outlined in the lesson is the following:
1) Find sample space.
2) Find number of successful outcomes.
3) Find Probability of Event = # of successful outcomes / sample space.

Thank you so much in advance for any help!

 

[MarkFL]

Hello and welcome to MHB, mdzvi! (Wave)

It's nice to see a parent taking an active hand in their child's education. (Yes) I will explain how I would go about answering the given question...

1.) The number of elements in the sample space is the total number $N$ of district debate team participants that can be formed by choosing 3 from the 10 members of the debate team. So, we use combinations here, since we are forming a subset:

\(\displaystyle N={10 \choose 3}=120\)

2.) To determine the number $S$ of successful outcomes, we need to look at the number $S_B$ of ways to choose 2 from the 6 boys while at the same time choosing 1 from the 4 girls, which we'll call $S_G$. We use combinations again, because we are choosing subsets:

\(\displaystyle S_B={6 \choose 2}=15\)

\(\displaystyle S_G={4 \choose 1}=4\)

Now, by the fundamental counting principle, we obtain:

\(\displaystyle S=S_B\cdot S_G=15\cdot4=60\)

3.) And so, if we call $X$ the event where 2 boys and 1 girl is chosen from the 6 boys and 4 girls, then we find the probability of $X$ happening as follows:

\(\displaystyle P(X)=\frac{S}{N}=\frac{60}{120}=\frac{1}{2}\)

 

[Evgeny.Makarov]

1.) The sample space is the total number $N$ of district debate team participants that can be formed by choosing 3 from the 10 members of the debate team.

More precisely, the sample space is the set of all combinations, i.e., subsets, of 3 students. The number of elements in the sample space is 120.

Edit: OK, the terminology may vary, especially since the OP wrote "# of successful outcomes / sample space", but the space is a set rather than a number in the official definition of probability space.

 

[MarkFL]

More precisely, the sample space is the set of all combinations, i.e., subsets, of 3 students. The number of elements in the sample space is 120.

Edit: OK, the terminology may vary, especially since the OP wrote "# of successful outcomes / sample space", but the space is a set rather than a number in the official definition of probability space.

Yes, upon reading your post, I do recall now that the sample space is actually a set, whereas I was calling the cardinality of the sample space the sample space itself. Thank you for the clarification! (Yes)

 

[mdzvi]

Thank you SOOO much for the replies. Now that I see the explanation, it makes perfect sense...and I feel silly we didn't understand it before.

As I said before, this is the last unit for his Geometry class, so it's just an introduction to Probability and Statistics. Basic. Learn the formula for calculating a Permutation. Learn the formula for calculating a Combination.

I think what threw us was that this was the ONLY problem where numerator of the Probability of an Event was the product of two Combinations. Every other problem only had 1 calculation for the numerator.

Thank you again for the help! Really, so appreciated. Have a great day!