7 girls and 6 boys go into a store at the same time. There are 5 salesmen to help them. Find the probability that the salesmen will first help (continued below)

Question

numbers of girls and boys.

 

1. P(3 girls or 3 boys)  ans. 105/143

 

2. P(at least 3 girls)  ans. 84/143

 

3. P(at least 2 boys) ans. 32/39

 

I have the correct answers for these Mutually Exclusive events in Adding Probabilities, but I am a substitute teacher, my degree is in music.  I need to help the students to use the correct process to arrive at the answers.  

 

Thank you in advance for any help you can give me.

Answer 1

First, it will be useful to figure out how many different ways there are for the salesmen to choose the first kids to help. There are 5 salesmen and a total of 7+6 = 13 kids. The salesmen will basically begin helping the kids at the same time, so "being helped first" just means "being one of the first 5." This means that it doesn't matter what order the first 5 kids are chosen: Sarah, Paul, Andrew, Jason, and Angela being chosen is the same as Paul, Angela, Jason, Andrew, and Sarah being chosen. When order doesn't matter, we can model the situation using combinations (as opposed to permutations, where order does matter).

 

So, how many different ways are there to choose 5 kids from 13?

 

"13 choose 5" = ₁₃C₅ = (13!) / ((5!)(13-5)!) = (13×12×11×10×9×8×7×6×5×4×3×2×1) / ((5×4×3×2×1)(8!)) = (13×12×11×10×9×8×7×6×5×4×3×2×1) / ((5×4×3×2×1)(8×7×6×5×4×3×2×1)) = (13×12×11×10×9) / (5×4×3×2×1) = 154440/120 = 1287 ways

 

(That looks complicated, but your students should have run into combinations by now and should be fairly comfortable with the formula, or at least with plugging it into their calculators.)

 

 

 

That's an important step. Now that we know how many possible ways there are to help the first kids, we'll just need to figure out how many of those 1287 total combinations fit each criterion above. If, for example, there were only 20 ways to see at least 2 boys (this is definitely not true), the probability of seeing at least 2 boys would be (successful combinations)/(total combinations) = 20/1287.

 

 

So, let's do that for each probability.

 

 

 

(1) How many combinations of 5 include exactly 3 girls or exactly 3 boys? These events are mutually exclusive (since we can't have both 3 girls and 3 boys), so we can calculate their probabilities separately and add them together.

 

 

Probability of exactly 3 girls:

 

For this situation, there will be exactly 3 girls and 2 boys. So, choose 3 of the 7 girls and 2 of the 6 boys:

 

(7 choose 3)×(6 choose 2) = (35)(15) = 525 ways to get exactly 3 girls

 

That means there are 525 out of the 1287 combinations that give you exactly 3 girls. This probability is 525/1287.

 

 

Probability of exactly 3 boys:

 

For this situation, there will be exactly 3 boys and 2 girls. So, choose 2 of the 7 girls and 3 of the 6 boys:

 

(7 choose 2)×(6 choose 3) = (21)(20) = 420 ways to get exactly 3 boys

 

That means there are 420 out of the 1287 combinations that give you exactly 3 boys. This probability is 420/1287.

 

 

Now, add those probabilities together, since the events are mutually exclusive:

 

525/1287 + 420/1287 = 945/1287 = 105/143

 

 

 

(2) This time, we're interested in how many combinations include at least 3 girls. That means our combination could have either exactly 3 girls, exactly 4 girls, or exactly 5 girls. Each of those three events is mutually exclusive. So, calculate their individual probabilities and add them together:

 

 

Probability of exactly 3 girls:

 

(7 choose 3)×(6 choose 2)

 

We know this from before. It's 525/1287

 

 

Probability of exactly 4 girls:

 

(7 choose 4)×(6 choose 1) = (35)(6) = 210

 

This probability is 210/1287.

 

 

Probability of exactly 5 girls:

 

(7 choose 5)×(6 choose 0) = (21)(1) = 21

 

This probability is 21/1287.

 

 

Add those together to get the total probability:

 

525/1287 + 210/1287 + 21/1287 = 756/1287 = 84/105

 

 

 

(3) You can do this one like the other two, or you can do it a slightly different way. It seems like it might be a pain to calculate how many combinations involve exactly 2 boys, exactly 3 boys, exactly 4 boys, and exactly 5 boys, so let's just calculate how many combinations involve exactly 0 boys or exactly 1 boy and subtract those. We're finding the complement of the set we're actually interested in: all the combinations that wouldn't work.

 

 

Probability of exactly 0 boys:

 

(7 choose 5)×(6 choose 0)

 

We know this from before. It's 21/1287.

 

 

Probability of exactly 1 boy:

 

(7 choose 4)×(6 choose 1)

 

We know this from before. It's 210/1287.

 

 

The total probability of having less than 2 boys is 21/1287 + 210/1287 = 231/1287. Now that we've figured out how many combinations won't work, subtract from the total combinations to figure out how many will. (We can do this because "less than 2 boys" and "at least 2 boys" are mutually exclusive, and, together, make up the entire sample space of combinations.):

 

1287/1287 - 231/1287 = 1056/1287 = 32/39

 

 

 

Hopefully this makes sense! Let me know if anything needs clarification.