(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Just finished it. I have a sneaky suspicion that I screwed it all up, so here I am to have my suspicion confirmed or refuted.

Question 1 said something like: "A class of 45 consists of 16 female math majors, 14 male math majors, 7 female non-math majors, and 8 male non-math majors. A group of 4 is randomly selected from the class. (i) In how many ways can this group have the same number of males as women? (ii) In how many ways can this group have the same number of males and females, if also it must have the same number of math majors as non-math majors."

Question 2 said something like: "Suppose an exam will consist of 5 problems chosen from 38, and suppose Jon knows how to do 30 of the 38. If he needs at least 4 problems correct to pass, what's the probability that he passes?"

Question 3 said something like: "(i) What's the probability that a hand of bridge has at 1 king? (ii) What's the probability that a hand of bridge has at least 1 king and at least 1 queen?"

2. Relevant equations

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3. The attempt at a solution

Question 1:

There are 23 females and 22 males in the class. For a group of 4 with an equal number of men and women, we obviously have to choose 2 men and 2 women. This means we have (23 choose 2)*(22 choose 2) possible choices.

If, also, we need an equal number of math majors and non-math majors, then we'll have either

2 male math majors, 2 female non-math majors;

2 female math majors, 2 male non-math majors;

or

1 male math major, 1 male non-math major, 1 female math major, 1 female non-math major.

Those are disjoint events and their sum is

(14 choose 2)*(7 choose 2) + (16 choose 2)*(8 choose 2) + (16 choose 1)*(14 choose 1)*(8 choose 1)*(7 choose 1).

Question 2:

Consider the numbers of ways to make an exam he won't pass. That would mean an exam with 0 questions he knows, only 1 question he knows, only 2 questions he knows, or only 3 questions he knows. Those are disjoint events and their sum is

S = (8 choose 5) + (8 choose 4)*(30 choose 1) + (8 choose 3)*(30 choose 2) + (8 choose 2)*(30 choose 3).

The probability of such an exam appearing would be S / (38 choose 5). Thus the probability of him passing would be 1 - S / (38 choose 5).

Question 3:

A hand with at least 1 king will either have 1 king only, 2 kings only, 3 kings only, or 4 kings. Those are of course disjoint events and their sum is

S = (4 choose 1)*(48 choose 12) + (4 choose 2)*(48 choose 11) + (4 choose 3)*(48 choose 10) + (4 choose 4)*(48 choose 9).

Then the probability of such a hand is S / (52 choose 13).

If our hand needs at least 1 king and at least 1 queen, then consider the complement of this event, that is, consider the number of ways to choose 0 kings and 0 queens, at least 1 king but 0 queens, and at least 1 queen but 0 kings; these are disjoint events. (The number of ways to choose at least 1 king but 0 queens is obviously the same as the number of ways to choose at least 1 queen but 0 kings; hence the factor of 2 in the following sum.)

S = (44 choose 13) + 2[(4 choose 1)*(44 choose 12) + (4 choose 2)*(44 choose 11) + (4 choose 3)*(44 choose 10) + (4 choose 4)*(44 choose 9)].

Then our final answer is 1 - S / (52 choose 13).

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# Homework Help: How'd I do on my probability exam?

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