Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 30 points Q6: 30 points Total: 100 points
FALSE. E(X) = 0(1/2) + 1(1/4) + 2(1/4) = 3/4. E(X2) = 0(1/2) + 1(1/4) + 4(1/4) = 5/4. Var(X) = E(X2) - E(X)2 = (5/4) - (3/4)2 = 11/16. It is not true that 11/16 is exactly three times 3/4.
FALSE. The variance of the sum of n independent random variables is the sum of the n individual variances, which in this case is 11n/16. Thus the standard deviation of the sum is the square root of 11n/16. Even if we didn't know the exact value of the variance of X, we could say that the given statement is false because the standard deviation must be of the form c*sqrt(n) for some number c, and this can't possibly equal 3n/2 for all values of n.
FALSE. The most straightforward way to compute this probability is to count the sets of three cards of three different suits and divide by (52 choose 3). To count the former sets we pick three suits in (4 choose 3) = 4 ways and then pick one card of each suit in 133 = 2197 ways, for 8788 total. Since (52 choose 3) = 52*51*50/1*2*3 = 26*17*50 = 1300*17 = 22100, the probability is 6591/22100 and without a calculator we can see that this is very close to 0.4 and thus greater than 3/8. (The fraction reduces to 169/425, and 3/8 of 425 would be 3*53.125 = 159.375, less than 169.)
It's possible to show that this probability is greater than 3/8 with much less calculation. Consider the cards as a sequence (a perm) rather than a set. The first card is guaranteed to be of a new suit. The second card has a probability of 39/51 of being of a new suit, slightly more than 39/52 = 3/4. The third card, if the second was of a new suit, has a probability of 26/50 of being of a third suit, slightly more than 1/2. So the probability of all three events happening in sequence is 1 * (more than 3/4) * (more than 1/2) = more than 3/8.
FALSE. First the straightforward way to calculate the probability: the number of sets of cards with no spades is (39 choose 3), so the probability we want is 1 - (39 choose 3)/(52 choose 3) = 1 - 39*38*37/52*51*50 because the 1*2*3 denominators of the choose functions cancel. We can reduce this last fraction a bit by cancellation to 3*38*37/4*51*50 = 38*111/51*200 = 4218/10200. Since this is clearly greater than 1/4, the probability we want is less than 3/4.
Here's another method that illustrates the use of expected value in solving probability problems. One reason you might think that 3/4 is the right answer is each of the three cards has a 1/4 chance to be a spade. The correct conclusion to draw from this is that the expected number of spades in the hand, which we know is also equal to 0*Pr(S=0) + 1*Pr(S=1) + 2*Pr(S=2) + 3*Pr(S=3). This is bigger than Pr(S=1) + Pr(S=2) + Pr(S=3), which the probability we want, because Pr(S=2) and Pr(S=3) are both positive. So since our probability is less than something that is equal to 3/4, the statement is false.
By the way, it's my usual practice to select booleans for my true/false questions randomly and independently, with "true" and "false" being equally likely. So it's a coincidence that all four answers were "false".
The autobahn travel time Ta, now assumed to be a normal random variable, has mean 110 and standard variation sqrt(400) = 20. For Hans to miss the deadline, this variable must be more than two standard deviations above its mean, which happens with probability about 0.025. (We know that there is about a 95% probability that it is within 2σ of the mean, and for a normal variable the "left tail" and "right tail" have the same size. It's the right tail that has him missing the deadline.
The back-road travel time Tb is also a normal variable, with mean 130 and standard deviation sqrt(25) = 5 minutes. For Hans to miss the deadline on the back roads, this variable must be more than four standard deviations above its mean, which is much less likely than for Ta to be two standard deviations above its own mean.
The Markov Inequality says that if X is never negative and has mean μ, the probability that |X - μ| is greater than or equal to cμ is at most 1/c. This is true for any constant c, so we just have to find the relevant c in each case.
To miss the deadline on the autobahn, Ta must be greater than 150 = μa*(150/110), so we can take c to be 15/11 and apply the Markov Inequality to say that the probability of missing the dealine is at most pa = 11/15. If we imagine a Ta that is slightly greater than 150 with probability slightly less than 11/15 and 0 otherwise, and thus has mean 110, we see that this bound could be close to the actual case.
To miss the deadline on the back roads, Tb must be greater than 150 = &mub*(150/130), so in this case c is 15/13 and the bound on the probability is pb = 13/15. We don't get as good a bound in this case (and neither bound gives Hans much comfort) because we are not using the information about the variance. It is that information that tells us that the back-road route is more reliable.
The Chebyshev Inequality says that for any random variable X with mean μ and standard deviation σ, the probability that |X - μ| is greater than or equal to cσ is at most 1/c2, where c is any positive number. (I really think the best way to memorize this result is to remember how it is proved using the Markov Inequality on the random variable (X = μ)2.)
We already found the correct c's in part (a). For Ta, Chebyshev tells us that Pr(|Ta - 110| ≥ 40) ≤ qa = 1/4. For Tb, Chebyshev tells us that Pr(|Tb - 130| ≥ 20) ≤ qb = 1/16. Unlike part (a), this case does not let us conclude that the two "tails" are of equal size, so these are the best estimates we have for Pr(Ta ≥ 150) and Pr(Tb ≥ 150).
We can either use the Product Rule for probabilities to get (1/6)5 or use the Probability Rule to get 1 (the number of ways to get five sixes) over 65 (the number of total ways to throw five six-sided dice). Multiplying out either expression gives 1/7776.
Once again the denominator is 65 = 7776. For the numerator, we must count sequences with three of one number, one of a second number, and one of a third. There are six choices for the first number, five for the second, and four for the third, for 6*5*4 = 120 choices of numbers. Given the three numbers, there are (5 choose 3) = 10 ways to pick three of the five positions for the number that occurs three times. Then we can put the second number left of the third number -- if we use both orders we count every sequence twice instead of once. So we have 1200/7776 or about 15%.
The two dice we throw can come out as 1-1, 2-2, 3-3, 5-5, or 6-6. Of course if they came out 4-4 we would have five of a kind, not a full house. The probability is 5/62 = 5/36.
To get four of a kind, she needs one four and one non-four, and they can be in either order: 4-1, 4-2, 4-3, 4-5, 4-6, 1-4, 2-4, 3-4, 5-4, or 6-4. Ten four-of-a-kinds out of 62 total throws gives a probability of 10/36 = 5/18.
To get five of a kind, she needs two fours, which is exactly one of 62 possible throws, a probability of 1/36.
There are three ways she could finish with five fours:
These three events are pairwise disjoint so we add the probabilities: (36/36*36) + (10*6/36*36) + (25/36*36) = (36+60+25)/36*36 = 121/1296 or a bit less than 10%.
As we discussed in lecture on 23 October, this is an example of the Law of Total Probability. The most common error was to add in (1/36) instead of (25/36)*(1/36) for the third case, forgetting that the third case only happens if the second throw gets no fours.
Last modified 23 October 2009