Tag: probability

I received a couple followup questions about Problem 7 in the”HW8-Counting” WebWork set. We discussed this exercise in class on Wednesday, but I thought I would post my reply in case it helps explain the reasoning:

This version of the question has the following numbers:

A bag contains 7 red marbles, 5 white marbles, and 9 blue marbles. You draw 3 marbles out at random, without replacement.

You are asked to compute 3 different probabilities:

What is the probability that all the marbles are red?

As usual, you have to compute how many outcomes are in the given event (in this case, the event that all the chosen 3 marbles are red), and divide by how many outcomes are in the sample space.

For this 1st part, you can compute in terms of permutations: there are P(7,3)=7*6*5 different permutations of choosing 3 red marbles (from the 7 in the bag), and P(21,3)=21*20*19 different permutations of choosing 3 marbles at random (from the 7+5+9=21 total marbles in the bag).

Hence, the probability that all 3 chosen marbles are red is

P(7,3)/P(21,3)=7*6*5/21*20*19

The next part asks:

What is the probability that exactly two of the marbles are red?

This part is trickier. I think it’s easier to do the computations in terms of combinations instead of permutations:

If we don’t care about the order in which there are chosen, there are C(21,3)=(21*20*19)/(3*2*1) different combinations of 3 marbles chosen from the 21 (“21 choose 3”). This is the size of the sample space if we think in terms of combinations; so that will be the denominator for calculating the probability.

For the numerator, we need to figure out how many combinations there are of 2 red marbles and 1 non-red marble (to get exactly 2 red marbles). There are “7 choose 2” combinations  of 2 red marbles (chosen from the 7 red marbles in the bag) and “14 choose 1” choices for the 1 non-red ball. You need to multiply these two numbers to get the total number of ways to get 2 red marbles and 1 non-red marble, i.e., C(7,2)*C(14,1)=[(7*6)/(2*1)]*[14/1].

(Note that it should be obvious that C(14,1)=14, or indeed C(n,1)=n for any positive integer n: there are n different ways of choosing 1 object from a set of n object!)

So the probability that exactly 2 of the marbles are red (and hence 1 is non-red) is

[C(7,2)*C(14,1)]/C(21,3)

You can figure out the 3rd part using the same techniques:

What is the probability that none of the marbles are red?

 

PS: It looks like there some of the later exercises that most of you are still working through. I’ll try to post some hints about these in the comments below tomorrow (and we can also discuss them in class on Monday).

PPS: Here’s an outline of how to approach Problem 9:

A box contains 55 balls numbered from 1 to 55. If 6 balls are drawn with replacement, what is the probability that at least two of them have the same number?

The key here, which happens with some probability computations, is to think in terms of the complement of the event you’re being asked about: we are trying to calculate P(E), where the event E = “at least two of the 6 balls  drawn have the same number”; consider instead the complement of E:

E^C = “no two of the 6 balls drawn have the same number”, i.e., “all 6 numbers drawn are different from each other.”

It’s easier to calculate the probability of E^C: there are 55*54*53*52*51*50 different outcomes in E^C (we have 55 choices for the first number; since the 2nd number chosen must be different from the 1st, there are 54 remaining choices; the 3rd number must be different from the 1st two, so there are 53 remaining choices; and so on).

Note that since the balls are chose with replacement, there are 55 choices on each of the 6 draws, meaning there are 55^6 different outcomes in the sample space.  Hence,

P(E^C) = 55*54*53*52*51*50/(55^6)

Therefore, P(E) = 1 – 55*54*53*52*51*50/(55^6)