Compute the Probability - Probability - Exam, Exams of Probability and Statistics

This is the Exam of Probability which includes Density Function, Probability, Unbiased, Same Dice, Conditioning, Company, Construction Project etc. Key important points are: Compute the Probability, Interval, Density Function, Jointly Continuous, Marginal Probability, Conditional Probability, Probability, Correlation CoeCient, Number of Heads, Special Case

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2012/2013

Uploaded on 02/21/2013

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MATH 4510: Review for Final Exam
1. Let Ube a uniform random variable on the interval (1,3). Let X= ln(U). Find the probability
density function of Xand compute the probability P(X1).
We compute FX(x) = P(Xx) = P(lnUx) = P(Uex) = ex1
31for 0 xln 3.
So, the probability density function of Xis
f(x) = d
dxF(x) = (ex
20<x<ln 3
0 otherwise .
So, P(X1) = R1
0f(x)dx=ex
2
1
0=e1
2.
2. The random variables Xand Yare jointly continuous with joint probability density function
f(x, y) = cxy for 0 < x, y < 1 and f(x, y ) = 0 otherwise. Find the constant cand compute the
marginal probability density functions fXand fY. Find the conditional probability density
function of Xgiven that Y=y. Find P(X Y 0.5).
We have that 1 = R1
0R1
0cxy dxdy=c
4, so that c= 4. We compute:
fX(x) = Z1
0
4xydy= 2xfor 0 <x<1
fY(y) = Z1
0
4xydx= 2yfor 0 < y < 1, and
fX|Y(x|y) = f(x, y )
fY(y)=4xy
2y= 2xfor 0 <x<1.
Finally,
P(XY 0.5) = Z Zxy 0.5
4xy dA=Z0.5
0Z1
0
4xy dydx+Z1
0.5Z1
2x
0
4xy dydx
=Z0.5
0
2xdx+Z1
0.5
2x
4x2dx=x2
0.5
0+ln x
2
1
0.5=1
4+ln 2
2.
Aside: We see that f(x,y ) = fX(x)fY(y), so that Xand Yare independent, so we could have
instead computed fX|Yby noting that fX|Y(x|y) = fX(x) when Xand Yare independent.
3. A coin that lands heads with probability pis flipped ntimes. Let Xbe the number of heads
and Ybe the number of tails. Compute the correlation coefficient ρ(X, Y ).
We have that Xand Yare both binomial with parameters nand p, so if we let σ2=np(1p),
then var(X) = var(Y) = σ2. Also, X+Y=n, so
cov(X, Y ) = cov(X, n X) = cov(X, n)cov(X, X) = var(X) = σ2.
Then,
ρ(X, Y ) = cov(X, Y )
pvar(X) var(Y)=σ2
σ2=1.
pf2

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MATH 4510: Review for Final Exam

  1. Let U be a uniform random variable on the interval (1, 3). Let X = ln(U ). Find the probability density function of X and compute the probability P (X ≤ 1).

We compute FX (x) = P (X ≤ x) = P (ln U ≤ x) = P (U ≤ ex) = e

x− 1 3 − 1 for 0^ ≤^ x^ ≤^ ln 3. So, the probability density function of X is

f (x) = d dx

F (x) =

ex 2 0 < x <^ ln 3 0 otherwise

So, P (X ≤ 1) =

0 f^ (x)dx^ =^

ex 2

∣^1

0 =^

e− 1

  1. The random variables X and Y are jointly continuous with joint probability density function f (x, y) = cxy for 0 < x, y < 1 and f (x, y) = 0 otherwise. Find the constant c and compute the marginal probability density functions fX and fY. Find the conditional probability density function of X given that Y = y. Find P (XY ≤ 0 .5).

We have that 1 =

0

0 cxy^ dxdy^ =^

c 4 , so that^ c^ = 4. We compute:

fX (x) =

0

4 xydy = 2x for 0 < x < 1

fY (y) =

0

4 xydx = 2y for 0 < y < 1, and

fX|Y (x|y) =

f (x, y) fY (y)

4 xy 2 y

= 2x for 0 < x < 1.

Finally,

P (XY ≤ 0 .5) =

xy≤ 0. 5

4 xy dA =

0

0

4 xy dydx +

  1. 5

∫ (^21) x

0

4 xy dydx

0

2 x dx +

  1. 5

2 x 4 x^2

dx = x^2

∣^0.^5

ln x 2

∣^1

ln 2 2

Aside: We see that f (x, y) = fX (x)fY (y), so that X and Y are independent, so we could have instead computed fX|Y by noting that fX|Y (x|y) = fX (x) when X and Y are independent.

  1. A coin that lands heads with probability p is flipped n times. Let X be the number of heads and Y be the number of tails. Compute the correlation coefficient ρ(X, Y ).

We have that X and Y are both binomial with parameters n and p, so if we let σ^2 = np(1 − p), then var(X) = var(Y ) = σ^2. Also, X + Y = n, so

cov(X, Y ) = cov(X, n − X) = cov(X, n) − cov(X, X) = − var(X) = −σ^2.

Then, ρ(X, Y ) = cov(X, Y ) √ var(X) var(Y )

−σ^2 σ^2

Note that this is a special case of the fact that ρ(X, Y ) = ±1 if and only if Y = aX + b for some real numbers a and b with a 6 = 0.

  1. A fair die is rolled 7 times. Compute the expected number of different sides that the die will land on.

Let Xi = 1 if the die lands on side i at least once and 0 otherwise. Let X be the num- ber of different sides that the die lands on, so that X = X 1 + · · · + X 6. We compute

E(Xi) = P (Xi = 1) = 1 − P (no roll lands on i) = 1 −

Then, E(X) =

i=1 E(Xi) = 6

6

  1. Alex plans to complete ten tasks. She expects that she will spend an average of 12 minutes doing each task with a standard deviation of 4 minutes, independently of how long she takes on other tasks. Approximate the probability that it will take her no longer than 150 minutes to complete all of the tasks.

Let Xi be the amount of time that Alex spends on the ith task and X = X 1 + · · · + X 10 be the amount of time that Alex spends on all of the tasks combined. Then, by the Central Limit Theorem,

P (X ≤ 150) = P

X 1 + · · · + X 10 − 10 · 12

≈ P (Z ≤ 2 .37) = Φ(2.37) = 0. 9911.