# Math 431 An Introduction to Probability. Final Exam Solutions

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1 Math 43 An Introduction to Probability Final Eam Solutions. A continuous random variable X has cdf a for 0, F () = for 0 < <, b for. (a) Determine the constants a and b. (b) Find the pdf of X. Be sure to give a formula for f X () that is valid for all. (c) Calculate the epected value of X. (d) Calculate the standard deviation of X. (a) We must have a = lim = 0 and b = lim + =, since F is a cdf. (b) For all 0 or, F is differentiable at, so f() = F () = if 0 < <, (One could also use any f that agrees with this definition for all 0 or.) (c) E(X) = f() d = 0 d = 3. (d) E(X ) = 0 d =, Var(X) = E(X ) [E(X)] = ( 3 ) = , 8 and σ(x) = Var(X) = =

2 . Suppose the number of children born in Ruritania each day is a binomial random variable with mean 000 and variance 00. Assume that the number of children born on any particular day is independent of the numbers of children born on all other days. What is the probability that on at least one day this year, fewer than 975 children will be born in Ruritania? We approimate the number of children born each day by a normal random variable. Letting X denote the number of children born on some specified day, and Z denote a standard normal, we have P (X 975) = P (X 974.5) = P ( X =.55) = P (Z +.55) Since each such random variable X (one for 0 each day) is assumed independent of the others, the probability that 975 or more children will be born on every day of this year is , and the probability that, on at least one day this year, fewer than 975 children will be born is close to %. 3. Suppose that the time until the net telemarketer calls my home is distributed as an eponential random variable. If the chance of my getting such a call during the net hour is.5, what is the chance that I ll get such a call during the net two hours? First solution: Letting λ denote the rate of this eponential random variable X, we have.5 = F X () = e λ, so λ = ln and F X () = e λ = (e λ ) = (.5) =.75. Second solution: We have P (X ) = P (X ) + P ( < X ). The first term is.5, and the second can be written as P (X > and X ) = P (X > )P (X X > ). The first of these factors equals P (X ) =.5 =.5, and the second (by virtue of the memorylessness of the eponential random variable) equals P (X ) =.5. So P (X ) =.5 + (.5)(.5) = Suppose X is uniform on the interval from to. Compute the pdf and epected value of the random variable Y = /X. We have if < <, f X () = Putting g(t) = /t we have Y = g(x); since g is monotone on the range of X with inverse function g (y) =, Theorem 7. tells us that y f Y (y) = d = if < y <, dy y y (Check: f Y (y) dy = / dy =.) We have E(Y ) = y y f Y (y) dy = / dy = ln. y (Check: E(/X) = f X() d = d = ln.) 5. I toss 3 fair coins, and then re-toss all the ones that come up tails. Let X denote the number of coins that come up heads on the first toss, and let Y denote the number of re-tossed coins that come up heads on the second toss. (Hence 0 X 3 and 0 Y 3 X.)

3 (a) Determine the joint pmf of X and Y, and use it to calculate E(X + Y ). (b) Derive a formula for E(Y X) and use it to compute E(X + Y ) as E(E(X + Y X)). (a) P (X = j, Y = k) equals P (X = j)p (Y = k X = j) = ( ) ) 3 j ( )j ( )3 j( 3 j k ( )k ( )3 j k = ( ) ) 3 j ( )3( 3 j k ( )3 j = ( )( ) 3 3 j j k ( )6 j whenever 0 j 3 and 0 k 3 j (and equals zero otherwise), so the joint pmf f = f X,Y has the following values: f(0, 0) = 64, f(0, ) = 3 64, f(0, ) = 3 64, f(0, 3) = 64, f(, 0) = 3 3, f(, ) = 6 3, f(, ) = 3 3, f(, 0) = 3 6, f(, ) = 3 6, f(3, 0) = 8. Hence E(X + Y ) = 0 + ( ) + ( ) + 3 ( ) = (Alternatively: X + Y is the total number of coins that come up heads on the first toss or, failing that, heads on the re-toss. Each of the three coins has a 3 chance of 4 contributing to this total, so by linearity of epectation, the epected value of the total is = (b) For each fied (0 3), when we condition on the event X =, Y is just a binomial random variable with p = and n = 3, and therefore with epected value pn = (3 ). Hence E(Y X) = (3 X) and E(X +Y ) = E(E(X +Y X)) = E(E(X X)+E(Y X)) = E(X + (3 X)) = E( X + 3) = E(X)+ 3 = = Let the continuous random variables X, Y have joint distribution f X,Y (, y) = (a) Compute E(X) and E(Y ). / if 0 < y < <, (b) Compute the conditional pdf of Y given X =, for all 0 < <. (c) Compute E(Y X = ) for all 0 < <, and use this to check your answers to part (a). (d) Compute Cov(X, Y ). (a) E(X) = f X,Y (, y) dy d = 0 0 dy d = 0 d =. E(Y ) = y f X,Y (, y) dy d = 0 0 y dy d = 0 d =. 4

4 (b) The marginal pdf for X is f X () = f X,Y (, y) dy, which equals 0 dy = for 0 < < (and equals zero otherwise). That is, X is uniform on the interval from 0 to. Hence for each 0 < <, the conditional pdf for Y given X = is f Y X (y ) = f X,Y (, y)/f X (), which is for 0 < y < and (c) E(Y X = ) = y f Y X(y ) dy = y 0 dy =. (We can also derive this answer from the fact that the conditional distribution of Y given X = was shown in (b) to be uniform on the interval (0, ), and from the fact that the epected value of a random variable that is uniform on an interval is just the midpoint of the interval.) To check the formula for E(Y X), we re-calculate E(Y ) = E(E(Y X)) = E( X) = E(X), which agrees with E(X) =, E(Y ) =. 4 (d) E(XY ) = 0 0 y dy d = 0 d =, so Cov(X, Y ) = E(XY ) E(X)E(Y ) = 6 ( )( ) = I repeatedly roll a fair die. If it comes up 6, I instantly win (and stop playing); if it comes up k, for any k between and 5, I wait k minutes and then roll again. What is the epected elapsed time from when I start rolling until I win? (Note: If I win on my first roll, the elapsed time is zero.) Let T denote the (random) duration of the game, and let X be the result of the first roll. Then E(T ) = E(E(T W )) = (E(T W = ) + E(T W = ) E(T W = 6 5) + E(T W = 6)) = ((E(T ) + ) + (E(T ) + ) (E(T ) + 5) + 0) = (5E(T ) + 5), 6 6 so 6E(T ) = 5E(T ) + 5 and E(T ) = 5 (minutes). 8. Suppose that the number of students who enroll in Math 43 each fall is known (or believed) to be a random variable with epected value 90. It does not appear to be normal, so we cannot use the Central Limit Theorem. (a) If we insist on being 90% certain that there will be no more than 35 students in each section, should UW continue to offer just three sections of Math 43 each fall, or would our level of aversion to the risk of overcrowding dictate that we create a fourth section? (b) Repeat part (a) under the additional assumption that the variance in the enrollment level is known to be 0 (with no other additional assumptions). (a) Since we do not know the variance, the best we can do is use Markov s inequality: P (X 06) 90.85; this is much bigger than.0, so to be on the safe side we 06 should create a fourth section. (b) Here we know the variance, but since normality is not assumed, we cannot use the Central Limit Theorem; we should use a two-sided or (better still) a one-sided

5 Chebyshev inequality. The two-sided inequality gives us P (X 06) P ( X 90 6) σ = <.0, so we re on the safe side with just three classes. (Or 6 56 we could use the one-sided Chebyshev inequality: P (X 06) P (X 90 6) σ = 0.07.) σ (a) A coin is tossed 50 times. Use the Central Limit Theorem (applied to a binomial random variable) to estimate the probability that fewer than 0 of those tosses come up heads. (b) A coin is tossed until it comes up heads for the 0th time. Use the Central Limit Theorem (applied to a negative binomial random variable) to estimate the probability that more than 50 tosses are needed. (c) Compare your answers from parts (a) and (b). Why are they close but not eactly equal? (a) The number of tosses that come up heads is a binomial random variable, which can be written as a sum of 50 independent indicator random variables. Since 50 is a reasonably large number, it makes sense to use the Central Limit Theorem, and to approimate X (the number of heads in 50 tosses) by a Gaussian with mean np = 50 = 5 and variance np( p) = 50 =.5. So P (X < 0) = P (X 9.5) = P ( X = ) = P (Z ), where Z is a standard Gaussian; using , we have P (X < 0) Φ(.56) 6%. (b) The waiting time until the 0th heads-toss is a negative binomial random variable, which can be written as a sum of 0 independent geometric random variables. 0 is a decent-sized number, so, as in part (a), we may apply the Central Limit Theorem and approimate W (the number of tosses required to get heads 0 times) by a Gaussian with mean r p = 0 / = and variance r( p) p = 0(/) 50.5) = P ( W 50.5 = 0.5 ) = P (Z 0.5 =. So P (W > 50) = P (W (/).66) Φ(.66) 5%. (c) Suppose the coin is tossed until it has been tossed at least 50 times and heads has come up at least 0 times. Then the outcomes for which X < 0 are precisely those for which W > 50, so the two events have equal probability. The reason we did not get the eact same answers in parts (a) and (b) is that the Central Limit Theorem is only an approimation, and when specific numbers are used there is likely to be some error.

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