004 444

A

333 33

3

B

2 22266

C

11 15

55

DEfron’s Dice

A beats B with probability 2/3.B beats C with probability 2/3.C beats D with probability 2/3.D beats A with probability 2/3.

A certain type of aluminum foil has an average of 1 flaw every 2 feet. What is the probability that there are at most 6 flaws in a given 10 feet of foil.

You may assume that the distribution of the flaws is a Poisson distribution.

Since X is assumed to be Poisson, it must be that λ = 5.

Let X denote the number of flaws in a 10-foot strip of foil.

We would expect, on average, to see 5 flaws in 10 feet.

So we have,P(X ≤ 6) = e−5 5k

k!k=0

6

∑ ≈ 0.762

If X is Poisson with 3P X = 1( ) = P X = 2( )

Then what is the value of P X = 4( )?

3e−λλ = e−λλ 2

2

λ = 6

P X = 4( ) = e−6 64

4!= 54e6

Suppose that a box contains 50 envelopes, 10 contain $8 and the others are empty. You choose 5 of the envelopes and will receive a reward of $3 plus the total amount of money contained in the 5 envelopes.

Let Y denote the amount of money you receive. Find the mean, variance and standard deviation for Y.

Let X denote the hypergeometric random variable with g = 10, N = 50, and n = 5.

Then Y = 8X + 3µY = E Y( ) = 8E X( ) + 3= 11

E X( ) = n gN

= 5 × 1050

= 1.0

Var X( ) = n ⋅ gN⋅ N − gN

⋅ N − nN −1

= 5 × 15× 45× 4549

= 0.7347

σ 2Y = 64σ

2X = 64 × 0.735 = 47.02

If Y = aX + bE(Y ) = aE(X)+ bVar Y( ) = a2Var(X)

Suppose that a box contains 500 apples, 200 of which are rotten. What is the probability that a sample of 10 apples contains exactly 4 bad apples?

If the sample is taken without replacement?

Let X denote the number of bad apples in the sample of 10.Then X is a hypergeometric random variable.And so,

�

P(X = 4) =

2004

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 3006

⎛ ⎝ ⎜

⎞ ⎠ ⎟

50010

⎛ ⎝ ⎜

⎞ ⎠ ⎟

≈ 0.2534≈ 0.2534

Suppose that a box contains 500 apples, 200 of which are rotten. What is the probability that a sample of 10 apples contains exactly 4 bad apples?

If the sample is taken without replacement? ≈ 0.2534

If the sample is taken with replacement?

Let X denote the number of bad apples in the sample of 10.Then X is a binomial random variable. In fact, X is b(n, p).

�

P(X = 4) =104

⎛ ⎝ ⎜

⎞ ⎠ ⎟ (0.4)4(0.6)6 ≈ 0.2508

Note that the binomial is a good approximation to the hypergeometric in this case.

If X is a random variable then the moment of X iskth E Xk( )So, the first moment is E(X), the mean.The second moment is E(X 2 ), used for finding the variance.

The moment generating function for X isMX (t) = E etX( )

provided for t in an (-h, h) for some h > 0.E(etX ) < ∞

MX 0( ) = 1′MX 0( ) = E X( )′′MX 0( ) = E X 2( )

MX(k ) 0( ) = E Xk( )

MX t( ) = E etX( ) = 1+ tM1 +t 2

2M 2 +

t 3

6M 3 +

t k

k!Mk +

If X is a discrete random variable and Y = f (X), then E Y( ) = f (k)p(k)k∑

The moment generating function for X isMX (t) = E etX( )

MX t( ) = E etX( ) = 1+ tM1 +t 2

2M 2 +

t 3

6M 3 +

t k

k!Mk +

If X is a discrete random variable and Y = f (X), then E Y( ) = f (k)p(k)k∑

′MX t( ) = ketk p(k)k∑

′′MX t( ) = k2etk p(k)k∑

MX(r ) t( ) = kretk p(k)

k∑

′MX 0( ) = kp(k) = E X( )k∑

′′MX 0( ) = k2p(k) = E X 2( )k∑

MX(r ) 0( ) = kr p(k)

k∑ = E Xr( )

MX t( ) = etk p(k)k∑ MX 0( ) = p(k)

k∑ = 1

What is the moment generating function for a b(n, p) binomial random variable X.

ektnk

⎛⎝⎜

⎞⎠⎟pkqn−k

k=0

∞

∑ =MX t( ) = ekt p X = k( )k=0

∞

∑ =

nk

⎛⎝⎜

⎞⎠⎟ekt pkqn−k

k=0

∞

∑ =nk

⎛⎝⎜

⎞⎠⎟et p( )k qn−k

k=0

∞

∑ = pet + q( )n

nk

⎛⎝⎜

⎞⎠⎟xk

k=0

n

∑ yn−k = x + y( )n

What is the moment generating function for a b(n, p) binomial random variable X.

MX t( ) = pet + q( )n

We will use the binomial’s moment generating function to get the mean and variance of a binomial random variable.

MX (t) = pet + q( )n Note that MX (0) = p + q( )n = 1

′MX (t) =ddt

pet + q( )n = npet pet + q( )n−1

µX = ′MX (0) = npe0 pe0 + q( )n−1 = np p + q( )n−1 = np

We will use the binomial’s moment generating function to get the mean and variance of a binomial random variable.

′′MX (t) =ddt

npet pet + q( )n−1⎡⎣

⎤⎦ =

= npet pet + q( )n−1 + n(n −1)p2e2t pet + q( )n−2

E X 2( ) = ′′MX (0) = np + n(n −1)p2

Var X( ) = E X 2( )− E X( )2 = np + n(n −1)p2 − np( )2

= np + n2p2 − np2 − np( )2 = np − np2 = np(1− p) = npq

What is the moment generating function for a geometric random variable? Let X be a geometric random variable with probability p of success on a given trial.

p(k) = P(X = k) = pqk−1

MX (t) = E etX( ) = ekt p(k) =k=1

∞

∑ ekt pqk−1 = p ektqk−1k=1

∞

∑k=1

∞

∑

= pq

ektqk =k=1

∞

∑ pq

qet( )k = pq⋅ qet

1− qet= pet

1− qetk=1

∞

∑

MX (t)

= pet

1− qet

What is the moment generating function for a geometric random variable? Let X be a geometric random variable with probability p of success on a given trial.

p(k) = P(X = k) = pqk−1

MX (t)=pet

1− qet

What is the moment generating function for a Poisson random variable? Let X be a Poisson random variable with probability p of success on a given trial.

MX (t)MX (t) = E etX( ) = ekt p(X = k)k=0

∞

∑ = ekt e−λλ k

k!k=0

∞

∑ =

k!λet( )k

e−λk=0

∞

∑ = e−λeλet

= eλ et−1( )