Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 20 points Q6: 40 points Total: 100 points
Question text is in black, solutions in blue.
The four true/false questions all involve a situation where two cards are drawn, without replacement, from a standard 52-card deck, such that each pair of cards is equally likely.
TRUE. The simplest argument is that after the first card is chosen, 3 of the other 51 cards have the same rank and 3/51 = 1/17. If you don't notice this trick, you can still get to the right answer fairly easily. There are 13 ranks and (4 choose 2) = 6 pairs of each rank, so 13*6 = 78 total pair hands. The total number of two-card hands is (52 choose 2) = 52*51/1*2 = 26*51. Dividing 13*6 by 26*51, canceling a 13, a 2, and a 3, gives 1/17.
TRUE. There are (52 choose 2) total hands -- how many contain one or more aces? There are (4 choose 2) with two aces. The ones with one ace each contain one of the four aces and one of the 48 other cards. So by the sum rule the total number of hands with aces is (4 choose 2) + 4*48 = 198 and by the probability rule the given answer is correct.
FALSE. Someone came up with quick argument -- once you pick the first card, there are eight cards that have rank adjacent to it but 12 of the same suit as it, so a flush is more likely, not less. The straightforward argument is that the number of straights is 13 (for choosing the rank sequence) times 4 times 4 (for choosing a suit for each rank) and 13*4*4 = 208, while the number of flushes is 4 (to choose the suit) times (13 choose 2) (to choose two cards of that suit) and 4*13*12/2 = 312. So there are more flushes, not fewer.
FALSE. We can answer this without computing the probabilities exactly, because clearly Pr(one or more aces) = Pr(one or more kings) by symmetry, and by Inclusion/Exclusion Pr(ace or king) = Pr(ace) + Pr(king) - Pr(ace and king). Since Pr(ace and king) is positive, Pr(ace or king) is strictly less than 2*Pr(ace), not greater. Since Pr(ace and king) = 4*4/(52 choose 2), we could compute that Pr(ace or king) = [198 + 198 - 16]/(52 choose 2), less than 2*[198/(52 choose 2)].
We need two tosses to get a match, so Pr(X = 1) = 0. Using 1 for heads and 0 for tails, we finish after two tosses in the cases 00 and 11, and continue if we have 01 or 10, so Pr(X = 2) = 2/4 = 1/2. For three tosses we finish after exactly three in the cases of 011 and 100, each of which has probability 1/8, so Pr(X = 3) = 2/8 = 1/4. For four tosses, we stop in the cases 0100 and 1011, so Pr(X = 4) = 2/16 = 1/8.
We observed above that Pr(X = 1) = 0. If n ≥ 2, there are exactly two ways to stop after n tosses -- we can have the last two results equal and all the earlier tosses alternating. (So this is 0101...011 or 1010...100 with n odd and 0101...100 or 1010...011 with n even.) The chance of any particular result of n tosses is 1/2n, so Pr(X = n) =2/2n = 1/2n-1. Another way to get this result is to note that you get to the second toss with probability 1, and after getting to the n'th toss you have exactly a 1/2 chance of getting to the n+1'st toss.
By the definition of expected value, we get the sum for n from 2 to infinity
of n/2n-1, which is 2/2 + 3/4 + 4/8 + 5/16 + ... which can be
written (1 + 2/4 + 3/8 + 4/16 + ... ) + (0 + 1/4 + 1/8 + 1/16 + ...). If the
original sum is S, the second is 1 + S/2 + 1/2, and S = (S+3)/2 implies s = 3.
This is obviously similar to Game A in the text (the geometric distribution)
which was the number of successes before the first failure -- recall that
E(A) = 1. In this game, we have the first toss, then some number of tosses
opposite to the last one (successes) before the first that agrees with the
last one (failure). So we get 1 for the first toss, expected number 1 for the
successes, then the failure -- this totals to 3.
A third way to see this is that we have one toss to move from the start
position to the "waiting" position. Then when waiting we have a 1/2 chance of
counting one toss and stopping, and 1/2 chance of counting one and still
waiting. This gives up E(X) = 1 + E(W), E(W) = (1/2)(1) + (1/2)(1 + E(W)),
hence E(W) = 2 and E(X) = 3.
There are 108 possible serial numbers, all equally likely (First Counting Problem). The number with no repeats is 108, by the Second Counting Problem. The probability is thus (10*9*8*7*6*5*4*3)/108, which is 1(0.9)(0,8)(0.7)(0.6)(0.5)(0.4)(0.3) = 0.018144 or a little under 2%.
Fourth counting problem, with k (number of elements chosen) = 8, n (number of possible elements) = 10, giving (n - 1 + k choose k) = (17 choose 8) = 17*13*11*10 (after cancellation) = 24310.
Third counting problem, with n = 10, but k can be any number from 1 through 8. So we have (10 choose 1) + (10 choose 2) + ... + (10 choose 8). This is a full-credit answer, but if we want the exact number it's easier to compute 210 - (10 choose 0) - (10 choose 9) - (10 choose 10) = 1024 - 1 - 10 - 1 = 1012.
We need to choose two ranks, for which there are (10 choose 2) possibilities. Then we need to choose 4 of the eight positions for the higher-ranked digit, and there are (8 choose 4) possibilities for this. Thus there are 45*70 = 3150 four-four serial numbers, and the probability is 3150/108 = 0.00003150, about one in 32,000.
Inclusion/Exclusion is actually easier. The number of sequences with exactly
four 4's, for example, is (8 choose 4)1494.
The desired set is the union of (exactly four 0's), (exactly four 1's),...,
(exactly four 9's). This union is ten times (8 choose 4)94, minus
the double-counted sequences, which are exactly the sequences counted in
part (d). So we get 4592000 - 3150 = 4588850. There is no need to worry
about triple-counting, etc., because no sequence of length 8 can have exactly
four of more than two types of digit.
The other natural method is extending d to get:
Each digit is a Bernoulli trial, as it is independently a 4 with probability 1/10. So the expected value is np = 8/10, and the variance is np(1-p) = 72/100.
Last modified 14 March 2008