1. ## 3 Probability Problems

1. SOLVED!

2. A cage contains 4 black and 4 white mice that are available for a biology experiment. Three mice are selected at random. What is the probability that at least one is black and at least one is white?

3.A salesperson plans to visit Annapolis, Bloomington, Carmel, Detroit, and Elksville on his business trip. Since Annapolis is not in the Midwest, he wants to visit that city last. If his boss randomly selects the order of all five cities to be visited, what is the probability that Annapolis is last?
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I'm figuring for the 2nd one I'd use the compliment in an equation like 1-Pr[something]-Pr[something]

As for the other one...not a clue

Anyone mind walking me through the process of solving these questions?

Edit: Solved the first one

Thanks.

2. Originally Posted by Sakurazaki
1. SOLVED!

2. A cage contains 4 black and 4 white mice that are available for a biology experiment. Three mice are selected at random. What is the probability that at least one is black and at least one is white?

3.A salesperson plans to visit Annapolis, Bloomington, Carmel, Detroit, and Elksville on his business trip. Since Annapolis is not in the Midwest, he wants to visit that city last. If his boss randomly selects the order of all five cities to be visited, what is the probability that Annapolis is last?

-----

I'm figuring for the 2nd one I'd use the compliment in an equation like 1-Pr[something]-Pr[something]

As for the other one...not a clue

Anyone mind walking me through the process of solving these questions?

Edit: Solved the first one.

Thanks.
2. Let X be the number of black mice chosen. Calculate 1 - Pr(X = 0) - Pr(X = 3). (A tree diagram easily gives these probabilities. It can also be done by listing the eight possible outcomes and getting the probability of each. It can also be done using the Hypergeometric distribution.).

3. Many approaches are possible. The simplest is to note that there since there are 5 cities there's a 1 in 5 chance that A will be last ....

3. 2. The outcomes I got were 8: BBB, BBW, BWB, BWW, WBB, WBW, WWB, WWW.
THe ones that fit the criteria are BBW, BWB, BWW, WBB, WBW, WWB. So I tried adding the 1/8 prob of each since 6 of the outcomes will work, meaning that there is a 6/8 or 3/4 probability that it will draw at least one white and at least one black.

However it seems my answer is wrong. Any idea on what I did wrong?

4. Originally Posted by Sakurazaki
2. The outcomes I got were 8: BBB, BBW, BWB, BWW, WBB, WBW, WWB, WWW.
THe ones that fit the criteria are BBW, BWB, BWW, WBB, WBW, WWB. So I tried adding the 1/8 prob of each since 6 of the outcomes will work, meaning that there is a 6/8 or 3/4 probability that it will draw at least one white and at least one black.

However it seems my answer is wrong. Any idea on what I did wrong?
Use a tree diagram to get the probability of the eight outcomes (because a mouse does not get replaced once selected). You have assumed that all outcomes are equally likely. This is not the case and a tree diagram clearly shows this.

5. @Mr. Fantastic: Sorry about editing my post, I honestly didn't know that was in the rules.

And I got .85 as my answer. It should be correct if I'm not mistaken.

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### a cage contains 4 black and 7 white mice that are available for a biology experiment. three mice are selected at random. what is the probability that at least one is black and at least one is white?

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