Probability Exam #2 Homework Questions

Let X be an exponentially distributed random variable, X = Exp (3). Find the pdf of Y = ln X

if h(x) = ln x, inverse of h(x) is e^x, so the density function of y is g(t) = f(e^t)e^t, which is 3e^t*e^(-3e^t)

Jobs arrive at a computer so that the time T between two consecutive jobs has an exponential distribution with mean 10 seconds. Find P(T< 5)

10 * integral from 0 to 5 of the pdf (e^-0.1t dt) = 1-e^-0.5t

Suppose that 100 birds are arranged in a circle. At a given moment each bird turns at random to the left or to right. Find the expected number of unpecked birds.

Ak is the event that a bird isn't pecked by either of its neighbors E[IAk]= P(Ak) = .5*.5 = .25 N = IA1 +...+ IA100 E[N] = E[IA1] +...+ E[IA100] =100*P(Ak) = 25

Consider the random point (X,Y) uniformly distributed in the triangle x, y>0, x + y < 1. What is the joint pdf? What is the marginal density of X? What is the conditional probability of Y? What is the expectation of the conditional probability?

Area is 1/2, so p(x,y) is 1/area, p(x,y) = 2 if within triangle, 0 otherwise Marginal of X is integral of the pdf with respect to y Conditional probability of Y is the joint probability / marginal probability...Unif(0,1-x) E[Y|X = x] = E[Unif(0,1-x)] = 1-x/2

A electronics store owner figures that 45% of the customers entering his store will purchase computers, 15% percent will purchase a smart TV set, and 40% will just be browsing. If 5 customers enter his store on a given day, what is the probability that he will sell exactly 2 computers and 1 smart TV set on that day?

C is the number of computer shoppers, T is the number of TV shoppers, and B is the number of browsing shoppers. C + T + B = 5, so P(C = 2, T = 1) = (5 choose 20) (0.45 ^2) (3 choose 1)(0.15)(0.4)^2

You bid on an object at a silent auction. You know you can sell it later for 100 and you estimate that the maximum bid from others is uniform on [70,130], which is continuous. How much should you bid to maximize your expected profit and what is the maximum expected profit?

Make a profit function: 0 if the max bid is greater than your bid, and 100-b if your bid is the greatest...expected profit is 1/interval (130-70 = 60) * integral from 70 to 130 of the profit function Maximum profit is the top bid (130)- the derivative of the expected profit function set equal to zero

Phone calls arrive to a call center such that the number of phone calls in a minute has a Poisson distribution with mean 4...lambda = 4. Conditional on the total number of callers, a caller is female with probability 0.5. In a given minute, let X be the number of female callers, Y the total number of male callers callers and N the total number of callers

N = X + Y ∼ Poi(4). P(X = i, Y = j), plug in i and j for x and y in the pmfs, multiply them together, they are independent!

A company consists of m = 24 men and w = 30 women. Each of the employee is to be randomly promoted with probability 1/3 independently of the other employees. Find the expected number of women that will be promoted given that the total number of employees promoted was 15.

N is the number of people promoted and W is the number of women promoted. Find P(W=k|N=15) = (30 choose k)*(24 choose 15-k)/(54 choose 15) P(W=k|N=5) = P(Y15 = k), Y15 = HGeom(30,24,15) E[W|N = 15] = E[Y15] = 15 · 30/54 ≈ 8.33

Suppose X is a standard normal random variable and Y is sigma*X + mu. The pdf of y is...

Normal variable with parameters of mu and sigma^2

If Y~N(mu, sigma^2), we have...

P(1/sigma * (Y-u) is less than or equal to mu + u*sigma -mu/sigma) = P (X is less than or equal to u) = Phi(u)

Jobs arrive at a computer so that the time T between two consecutive jobs has an exponential distribution with mean 10 seconds. Find P(T > E[T])

P(T > E[T]) = P(T > 10) = e^-10lambda = 1/e

A system consists of four components which function independently with probabilities 0.9, 0.8, 0.6, and 0.6. Let X denote the number of components that work. Find P(X = 1).

P(X=1)=P(X1 =1, X2 =X3 =X4 =0)+P(X2 =1, X1 =X3 =X4 =0) +P(X3 =1, X1 =X2 =X4 =0)+P(X4 =1,X1 =X2 =X3 =0) = 0.0512.

Let X = Exp (lambda) and Y = lambda * X. Show that Y = Exp (1)

P(lambda * X < x) = P(X < x/lambda) = lambda * integral of 0 to x/lambda e ^-lambda x dx = 1-e^-lambda (x/lambda) = 1-e^.x, which is the same as Exp(1)

Let (X,Y) be uniform on four points (0,0) (1,0) (1,1) (2,1). Find the marginal pmfs of X and Y. For which joint pmf of (X,Y) are X and Y uniform on their respective ranges?

Range of X is {0,1,2}, range of Y is {0,1}. If p is the joint pmf of (X,Y), px(0) = 1/4, px(2) = 1/4, px(1) = 1/2 py(0) = 1/2, py(1) = 1/2 p(0,0)=p(1,0)+p(1,1)=p(2,1)= 1/3 p(0,0)+p(1,0)=p(1,1)+p(2,1)= 1/2

Let us look again at the situation analyzed in Example 3.7, where N is the total number of callers, X the number of female callers and Y is the number of male callers at a call center. Y is independent of X and X, Y ∼ Poi(2).

Thus P(X =i,Y =j)=P(X =i)P(Y =j). The conditional pmf of N given X = i is pN|X=j(i+j)=P(N =i+j|X =j) = P(X = i and Y = j)/P(X = j) = P(Y = j) = e^-2(2^j/j!)

A company consists of m = 24 men and w = 30 women. Each of the employee is to be randomly promoted with probability 1/3 independently of the other employees. Find the expected number of women that will be promoted.

W ∼ Bin(30, 1/3) so E[W ] = 30/3 = 10.

Consider again the matching problem. Given the n drunken sailors picking randomly their hats, we denote by Nn the number of matches, i.e., the number of sailors that end up picking their own hat. We want to compute the expectation of Nn.

We denote by Hk the event "sailor k pick his own hat"...P(Hk) = 1/n We denote by IHk the indicator of Hk, i.e., the random variable that has value 1 if Hk occurs, and value 0 otherwise. We see that IHk is a Bernoulli random variable with winning probability 1 so E[IH] = P(Hk) = 1/n...Nn =IH1 +···+IHn = E[N] =E [I1] +···+E[In] = 1/n +···+ 1/n =1.

Let X and Y be independent and ∼ Unif[0, 1]. Find (a) E[XY ], (b) E[X/Y ], (c) E[ln(XY)], and (d) E |Y −X| .

a) E[XY] = E[X]E[Y] = (1/2)^2 b) E[X/Y] = E[X]E[1/Y] = 1/2E[1/Y] = 1/2 * the integral from 0 to 1 of 1/y dy = infinity c) ln[XY] = ln X + ln Y, so E[lnXY] = E[lnX] + E[lnY] = 2E[lnX] = 2 times the integral from 0 to 1 of log x dx d)E[X -Y] = double integral over [0,1]x [0,1] of x-y dx dy, which is split into two integrals (two triangles make up an absolute value graph), but both are equal so you can just 2 times one of the integrals

Let X and Y be independent and have the same geometric distribution with success probability p. Find the conditional distribution of X given X + Y = n. Explain intuitively.

joint pmf is p(j,k) = P(X = j, Y = k) = q^(j+k-2)p^2 X + Y ~ NegBin(2,p), so P(X+Y = n) = (n-1 choose 1)p^2q^n-2 P(X = j, | X+Y = n) = P(X=j, Y = n-j)/P(X+Y = n) = q^n-2*p^2/(n-1)p^2q^n-2 = 1/(n-1)

The pdf of a random variable is given by f(x) = c/sqrt(1-x^2) from -1<x<1. Find the value of c and the cdf of x.

integral from -1 to 1 of c/sqrt(1-x^2) dx = 1, take the antiderivative (which is sin^-1 (x)), pull c out, solve for c

An urn contains 999 balls labelled 1, . . . , 999. Draw at random a ball from the urn and let L denote its label. Next, roll a die L times and record the number of times N that you get a 6. We want to compute the expectation of N, i.e., the expected number of 6's we get.

p = 1/6, q = 5/6 (N|L = l)~Bin (l, p) E[N|L=l] = pl = l/6 E[N] =E[N |L = 1]P(L = 1) + E[N |L = 2]P(L = 2) + · · · + E[N |L = 999]P(L = 999) = (1/999)*(1/6) + (2/999)*(1/6)+...+(999/6*999) = 83.3333

Suppose that (X, Y ) is uniformly distributed in the unit disk. Then p is the uniform distribution on the unit disk. what is the pdf? What is the conditional pdf of Y given X = x0?

1/pi * 1 if within the disc, 0 outside the disc Y|X = x0 ~Unif (-sqrt(1-x^2), sqrt(1-x^2))

A system consists of four components which function independently with probabilities 0.9, 0.8, 0.6, and 0.6. Let X denote the number of components that work. Find P(X > 0).

= 1-P(X = 0) = 1- (0.1)(0.2)(0.3)(.4) = 0.9976

Suppose that 20 of birds labelled 1, . . . , 20 are sitting on a circle facing its center. At a given moment, each bird turns randomly and with equal probability to the left or right to see who is his neighbor. We denote by N the number of birds not seen by either of their neighbors. We want to compute the expectation of N . For k = 1, . . . , 20 we denote by Ak the event "the k-bird is not seen by either of its neighbors". We denote by IAk the indicator of Ak, i.e., the random variable that is equal to 1 if A has occurred and 0 otherwise.

E[IAk ] = P(Ak) = 0.5 · 0.5 = 0.25 N =IA1 +···+IA20. E[N ] = E[IA1 ] + · · · + E[IA20 ] = 20 · P(A1 ) = 5.

Suppose you perform another experiment: you roll a fair die until you get a 6. Record the number R of rolls, and then flip a fair coin R times, and record the number N of heads you get. Find the expectation E[N] of N.

E[N ] = E[N |R = 1]P(R = 1) + E[N |R = 2]P(R = 2) + · · · + E[N |R = m]P(R = m) + · · · . F is a geometric random variable with success probability p = 1/6, so P(R = m) = q^m-2 * p Number of heads is a binomial random variable with m trials and probability 1/2, so E[N|R = m] is m/2 E[N] = 1/2(p + 2pq + mpq^m-1) = 3

Suppose you are performing the following random experiment: you flip a fair coin until you get a head. Record the number F of flips it took, and then roll a fair die F times and record the number N of 6-s you obtain. Find the expectation E[N] of N.

E[N] = E[N]=E[N|F =1]P(F =1)+E[N|F =2]P(F =2)+···+E[N|F =m]P(F =m)+··· F is a geometric random variable with success probability 1/2, so P(F = m) = 1/2^m Number of 6s is a binomial random variable with m trials and success probability 1/2, so E[N|F=m] = m/6 E[N] =1/6*(1/2^1+...+m/2^m) = 1/3

Suppose that n man-woman couples go to a meeting. The individuals are randomly placed around a table. Denote by N the number of couples that have neighboring seats. Find E[N].

E[N] = n*P(Wn) = 2n/2n-1

Let X1, X2, . . . , Xn be i.i.d. random variables with mean μ and variance σ2, let Sn = X1 +···+Xn, and let X ̃ = Sn/n (called the sample mean). Find E X ̃ and var X ̃ .

E[Sn] = E[X1+...+Xn] = E[X1]+...+[Xn] = nE[X1] = n*mu...var = nsigma^2 E[X~] = E[Sn/n] = E[Sn]/n = mu Var[X~] = var[Sn/n] = var[Sn]/n^2 = sigma^2/n

Jobs arrive at a computer so that the time T between two consecutive jobs has an exponential distribution with mean 10 seconds. Find the variance.

E[T] = 10 = 1/lambda, so lambda = 0.1...variance = 1/lambda^2 = 100

Let U and V be independent and Unif[0,1] and let X = min(U,V) and Y = max(U, V ). Find cov[X, Y ] and comment on its sign.

E[X] + E[Y] = E[U] + E[ V] = 1 E[Y-X] = E[Y] - E[X] = E[abs(U-V)] = double from 0 to 1 and 0 to 1 of abs(U-V) dv du = 1/3... E[Y+X] =1 E[Y-X] =1/3 Plug into cov equation

Let x be the random variable with pdf f(x) = .5(e)^-(x) from negative infinity to infinity. Find the expectation and variance

Expectation: density function is odd (if you plug in a negative number, it isn't the same result) so expectation is 0 Variance: integral of negative infinity to infinity of x^2 * f(x), integrating by parts gives you 2

Flip a fair coin repeatedly and wait for the first occurrence of 3 consecutive heads, HHH. Find the expected number of flips until this occurs.

F is the number of coin flips until the first string of n consecutive heads. μn = E[Fn] =E[Fn|T =1]P(T =1)+E[Fn|T =2]P(T =2)+···+E[Fn|T =n]P(T =n) P(T= k) = 1/2^k...P(T>n) = 1/2^n E[Fn|T=k]=μn+k, ∀k=1,2,...,n, E[Fn|T>n]=n. mu(n) = 2^n(sum from 1 to n of k/2^k) + n

Two species of fish have weights that follow normal distributions. Species A has mean 20 and standard deviation 2; species B has mean 40 and standard deviation 8. Which is more extreme: a 24 pound A fish or a 48 pound B fish?

Fish A is 2 standard deviations away from the mean, while fish B is only one standard deviation away---so fish A is more unlikely and more extreme

A system consists of four components which function independently with probabilities 0.9, 0.8, 0.6, and 0.6. Let X denote the number of components that work. Find the expectation and variance.

For i = 1,2,3,4, denote Xi the random variable that is equal to 1 if the i component is working and 0 if it's not (they are all Bernoulli variables) X = X1+...+X4 E[X] = E[X1]+...+E[X4] = 0.9 + 0.8 + 2(0.6) = 3 var[X] = var[X1]+...+var[X4]

Jobs arrive at a computer so that the time T between two consecutive jobs has an exponential distribution with mean 10 seconds. Find the probability that the net job arrives within 5 seconds given that the last job arrived 25 seconds ago

P(T<30 given T > 25) = 1-P(T>30 given T > 25) = 1-P(T>5) = P(T<5)

A large number of lightbulbs are turned on in a new office building. A year later, 80% of them still function, and 2 years later, 30% of the original light bulbs still function. Does it seem like this follows an exponential distribution?

P(T>1) = e^-lambda... so .8 = e^-lambda P(T>2)=e^-2*lambda...so .3 = e^-2*lambda...not the same lambda

Choose a number X at random from the set of numbers {1, 2, 3, 4, 5}. Now choose a number at random from the subset of numbers ≤ X, that is, from {1, . . . , X}. Call this second number Y . Find the joint mass function of (X, Y) and the marginal pdf of Y. Are X and Y independent?

P(X = i, Y = j) = 1/5(i-1) for any i > j, 0 if i is less than or equal to j P(Y = j) = P(X = j +1, Y = j) + ...+ P(X = 5, Y = j) = 1/5 (1/j + ... + 1/4). They are dependent because the cmf depends on the value X = 1---P(y|x) =1/i-1

Suppose that X ∼ Bin(m, p) and Y ∼ Bin(n, p) are independent binomial random variables with the same probability of success p. We want to compute the conditional pmf of X given that X + Y = r, i.e., the function P(X = j|X + Y = r).

P(X = j|X + Y = r) = P(X = j and Y = r-j)/P(X+Y = r) X+Y ~Bin(m+n, p) P(X = j|X + Y = r) = (m C j)(n C r-j)/(n+m C r)

Suppose that n man-woman couples go to a meeting. The individuals are randomly placed around a table. For k = 1,2,... find the probability that the kth woman sits near her partner.

Wk is the event that the kth woman sits near her partner at the round table P(W1) = P(Wn)= 2/2n-1... n = number of couples

The joint probability mass function of (X, Y ) is given by p(1,1) = 1/8 , p(1,2) = 1/4, p(2,1) = 1/8, p(2,2) = 1/2. Compute the conditional mass function of X given Y = i, i = 1,2. Are X and Y independent?

X and Y have the same range, {1,2}. pX|Y (1, 1) = p(1, 1)/pY (1) = 1/2 pX|Y (2, 1) = p(2, 1)/pY (1) = 1/2 pX|Y (1, 2) = p(1, 2)/pY (2) = 1/3 pX|Y (2,2) = p(2,2)/pY (2) = 2/3 As pX|Y (1,1) does not equal pX(1) = 3/8 the random variables are not independent.

A stick measuring one yard in length is broken into two pieces at random. Compute the expected length of the longest piece.

X is the location where you break the stick, L is the function that represents the length of the longest of the two segments-- L(X) = 1-X if X < .5 L(X) = X if X > .5 E[L] = integral from 0 to .5 of (1-x) plus the integral of .5 to 1 of x = 3/4

A saleswoman working for a company sells goods worth X×$1000 per week, where X is Unif[0,2]. Of this, she must pay the company back up to $800 and gets to keep the rest. Compute her expected profit (i) in a given week, and (ii) in a week when she makes a profit.

Y is the profit...Y = 1000X-800 a) E[Y] = 1000E[X]-800 = 1000-800 = 200 b) E[Y|Y>0] = E[1000X-800|X>0.8] = 1000E[X|X>0.8]-8000 = 1000/P(X>0.8) * integral from 0.8 to 2 of xpx(x)dx = 600

Is it true in general that p(x,y) < p(x) for all x and y?

Yes because px(x) is the sum of nonnegative terms of p(x,y), and the "and" condition of x and y is contained with in p(x)

Let X and Y be independent and ∼ Exp(1). Find E e−(X+Y )/2 .

e^−(X+Y )/2 = e^−X/2*e^−Y/2 Because X and Y are independent, you can multiply their expectations together Because they are identically distributed, we have E[ −X/2]=E[−Y/2] is the integral from 0 to infinity of e^-x/2 * e^-x dx, which is the integral from 0 to infinity of e^-3x/2 dx, which is (2/3)^2

The element nobelium has a half life of 58 minutes. Let X be the lifetime of an individual nobelium atom. Find P (X > 30)

half-life = e^-lambda * half life = P(T>h) = 1/2...1/2 = h = ln2/lambda, lambda = ln2/h

Let f(x,y) = 24xy, x,y≥0, x+y≤1, 0, otherwise. (i) Show that f (x, y) is the joint pdf of a continuous random vector (X, Y ). (ii) Find the marginal pdf of X. (iii) Find E[X] and E[Y ]. (iv) FindE[Y|X=x],x∈(0,1).

i) Do the double integral over the region, it should equal 1! Double integral over the region: first from 0 to 1, second from 0 to x of 24xy dydx, solve, this should equal one! ii) fx(x) = integral of the region of the pdf with respect to y iii) E[X] = single integral of x* pdf with respect to x (use given bounds of x) E[Y] = E[X] because it is symmetrical iv) E[Y|X=x] = 1/fx(x) * integral of yf(x,y)dy, using given bounds of y...should be in terms of x!

Suppose that we draw 2 cards out of a regular deck of 52. We denote by x the number of Hearts drawn and by Y the number of Queens. There are 13 Hearts and 4 Queens, and exactly one of the Queens is the Queen of Hearts, so it counts as both. We cannot have 2 hearts and 2 queens. Find the joint pmf of (X,Y)

p(hearts, queens) p(0,0) = (36 C 2)/(52 C 2) p(0,1) = (3 C 1) * (36 C 1)/(52 C 2) p(0,2) = (3 C 2) / (52 C 2) p(1,0) = (12 C 1) * (36 C 1) / (52 C 2) p(2,0) = (12 C 2) / (52 C 2) p (1,2) = (3 C 2)/ (52 C 2) p(1,1) = (12 C 1) (3 C 1) + (36 C 1) /(52 C 1)

Consider a family with three children. Let X be the number of daughters and Y the number of sons. Find the joint pmf of (X,Y)

p(i,j) = P(X = i, Y = j), For i+j = 3... p(i,j) = (3 choose i) 1/2^3