Test 2 is rescheduled for the first hour or so of class on Monday 26 March.
Corrected answers:
MAT2572Test2ReviewAnswersSpring2018
Below the fold are worked-out solutions to the first few problems, similar to what I did in class. They use the basic counting techniques which are described in these notes: MAT2572CountingMethods-slideshow
Note: I have not simplified the answers in the solutions below. The simplified answers are on the answer sheet.
1a) We can do this by using the multiplication rule, because we can think of creating a 7-digit binary number by choosing the digits one at a time, and the number of choices at each step is 2 no matter what we have chosen previously. So the number of 7-digit binary numbers is
2*2*2*2*2*2*2 = 27.
1b) There are two ways to think about this problem.
One way is to think that we know that our number must contain three 1’s, and once we have picked out positions in the number for those three 1’s to be in, there is no more choice (the remaining four slots must be filled with 0’s). So that would mean that there are $\binom{7}{3}$ ways to make such a binary number.
Another way is to think of a 7-digit binary number with three 1’s in it as a permutation of the digits 1110000, where the 1’s and the 0’s are indistinguishable from each other. So the number of such permutations is 7! (the number of permutations of the 7 digits) divided by 3! (because we overcount by the number of permutations of the three indistinguishable 1’s) and also divided by 4! (because we overcount by the number of permutations of the indistinguishable 0’s). So this gives $\frac{7!}{3!4!}$, which is the same as the previous answer.
2a) Here “but” stands in for “and”: this is an intersection probability.
Let A = “the first card is an ace” and B = “the second card is not an ace”.
Then $P(A \cap B) = P(A)P(B|A)$
$P(A) = \frac{4}{52}$ because there are four aces int he deck, and we have is equally likely outcomes.
$P(B|A) = \frac{48}{51}$ because if the first card is an ace, there are still 48 non-aces in the deck, but there are only 51 cards total left in the deck.
So $P(A \cap B) = \frac{4}{52}\cdot \frac{48}{51}$
Another way: Using the multiplication rule, there are $4\cdot 48$ ways to get an ace and then a non-ace. There are $_{52}P_{2} = 52\cdot51$ total ways to choose two cards in sequence from the deck, so the probability is $\frac{4\cdot 48}{52\cdot 51}$
2b) There are $\binom{13}{2}$ ways to choose two diamonds, and there are $\binom{52}{2}$ ways to choose two cards from the full deck of 52 cards, so the probability is $\frac{\binom{13}{2}}{\binom{52}{2}}$
2c) (Note: this is closely related to a problem which appeared on a quiz! MAT2572Quiz7MarchSolution)
In this problem, the multiplication rule (whether of counting or of probability) cannot be used directly, because there is one card which is both a club and an ace. So we have to break the computation into two parts, depending on whether the first card was a non-ace club or the ace of clubs.
There are 12 non-ace clubs, and for each of them there are 4 choices of ace which could be the second card, so this gives 12*4 = 48 ways to get a club and then an ace.
If the first card is the ace of clubs, there are only 3 possibilities for the second card to be an ace, so that gives 3 more ways to get a club and then an ace.
So there is a total of 48+3 = 51 ways to get a club first and then an ace.
The total number of ways of choosing 2 cards, in sequence, from the deck is $_{52}P_{2} = 52\cdot 51$
So the probability of getting first a club and then an ace is $\frac{51}{52\cdot 51} = \frac{1}{52}$.
