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Questions 1.A full h ou s e in poker is a hand where th re e cards sh are one rank and two cards share another rank. How many ways are the re to g e t a full-house? W ha t is t h e probability of getting a full-house?

2.20 politicians are having a te a party, 6 Democrats and 14 Republicans. To prepare, t h e y ne ed to choose: 3 p e op l e to s e t t h e table, 2 p e op l e to boil t h e water, 6 p e o p l e to make t h e scones.

Ea c h p er son can only do 1 task. (N ote th at this doesn’t add u p to 20. T h e re st of t h e p e o p l e don’t he l p.) (a) In how many different ways can t h e y c h o os e which p e o p l e perform t h e s e t as ks ? (b)S u p p o s e th at t h e Democrats all h ate tea. If t h e y only give te a to 10 of t h e 20 peopl e, what is t h e probability th at t h e y only give te a to Republicans? (c)If t h e y only give te a to 10 of t h e 20 people , what is t h e probability th at t h e y give te a to 9 Republicans and 1 Democrat?

3.Let C and D b e two events with P ( C ) = 0.25, P ( D ) = 0.45, and P (C ∩ D ) = 0.1. W ha t is P (C c ∩ D ) ?

4.M ore cards! S u p p o s e you want to divide a 52 card dec k into four hands with 13 cards each. W h at is t h e probability th at e ac h hand h a s a king?

5.Corrupted b y their power, t h e j u dg e s running t h e popular g am e show America’s Next Top Mathematician have b e e n taking bribes from many of t h e contestants. Ea c h episode, a given contestant is either allowed to s tay on t h e show or is kicked off.

If t h e contestant h a s b e e n bribing t h e j u dg e s s h e will b e allowed to stay with probability 1. If t h e contestant h a s not b e e n bribing th e judges, s h e will b e allowed to stay with probability 1/3.

S u p p o s e th at 1/4 of t h e contestants have b e e n bribing t h e judges. T h e s am e contestants bribe t h e j u dg e s in b o t h rounds, i.e., if a contestant bribes t h e m in t h e first round, s h e bri bes t h e m in t h e se cond round too ( and vice versa). (a)If you pick a random contestant who was allowed to stay during t h e first episode, what is t h e probability tha t s h e was bribing t h e j u d g e s? (b)If you pick a random contestant, what is t h e probability th at s h e is allowed to stay during b o t h of t h e first two e p i s od e s ? (c)If you pick random contestant who was allowed to st ay during t h e first episode , w hat is t h e probability t ha t s h e g e t s kicked off during t h e se cond e p i s od e ?

6. Th e re is a screening t e s t for p rostate cancer th a t looks at t h e level of P S A ( prostate s pe c i fi c antigen) in t h e blood. Th e re are a number of reasons b e s i d e s prostate cancer th at a man can have elevated P S A levels. In addition, many t y p e s of prostate cancer

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to distinguish t h e different t y p e s and using t h e t e s t is controversial b e c a u s e it is hard to quantify t h e accuracy rates and t h e harm done b y false positives.

For this problem we’ll call a positive t e s t a true positive if it c at c h e s a dangerous t y p e of prostate cancer. We’ll as s u m e t h e following numbers: Rate of prostate cancer among men over 50 = 0.0005 True positive rate for t h e t e s t = 0.9 False positive rate for t h e t e s t = 0.01 Let T b e t h e event a man h a s a positive t e s t and let D b e t h e event a man h a s a dangerous t y p e of t h e disease. Find P (D|T ) and P (D|T c ).

7.

S u p p o s e now th at events A, B and C are mutually independent with P ( A ) = 0.3,

P ( B ) = 0.4, P ( C ) = 0.5.

Comp ute t h e following: (Hint: Use a Venn diagram) (i ) P ( A ∩ B ∩ C c ) (ii) P ( A ∩ B c ∩ C ) (iii) P (A c ∩ B ∩ C ) 8. S u p p o s e X is a random variable with E ( X ) = 5 and Va r ( X ) = 2. W h at is E ( X 2 ) ?

9.S u p p o s e I play a gambling g am e whe re I either win or lose k dollars. S u p p o s e further th at t h e chance of winning is p = .5.

I employ t h e following strate gy to try to guarantee th at I win s om e money.

I b e t $1; if I lose, I double my b e t to $2, if I lose I double my b e t again. I continue until I win. Eventually I’m sure to win a b e t and net $1 (run through t h e first few rounds and you’ll s e e why this is t h e net).

If this really worked casinos would b e out of business. Our goal in this problem is to understand t h e flaw in t h e strategy. (a)Let X b e t h e amount of money b e t on t h e last ga m e ( t h e one I win). X take s values 1, 2, 4, 8, . . . . Determine t h e probability m a s s function for X . T h a t is, find p(2 k ), whe re k 10.in {0, (a)1,S2, u p. p. o. s}.e th at X h a s probability density function f X ( x ) = λe − λ x for x ≥ 0. is (b) C om pute E (Comp X ) . ute t h e cdf, F X ( x ) . (c) Use your answer ) to explain t h e cdf st aof te g (b) If Yin=part X 2 , (cbom p ut e t h e pwhy df and Y.y is a b ad one.

11.

Let X b e a discrete random variable with pm f p given by: x

−2

−1

0 2 p ( x ) 1/15 2/15 3/15 4/15 5/15

(a) Le t Y = X 2 . Find t h e pm f of Y . (b) Find t h e value t h e cdf of X at -1/2, 3/4, 7/8, 1, 1.5, 5.

1

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Live Exam Helper 12. For e ach of t h e following say w he the r it can b e t h e g rap h of a cdf. If it can b e, say wh ethe r th e variable is discrete or continuous. (i) (ii) (iii) F (x) F (x) 1 1 1 F (x) 0.5 0.5 0.5 x x x (iv) 1 0.5

( v)

F (x)

1 0.5

x

(vi)

F (x) x

1 0.5

F (x)

x (vii) 1 0.5

(viii)

F (x) x

1 0.5

F (x) x

13. Normal Distribution: Throughout t h e s e problems, let φ and Φ b e t h e p df and cdf, respectively, of t h e standard normal distribution S u p p o s e Z is a standard normal random variable and let X = 3Z + 1. (a) E xp re s s P ( X ≤ x ) in te rm s of Φ (b) Differentiate t h e expression from ( a ) with re s p e c t to x to g e t t h e p d f of X , f ( x ) .

Re m em be r t ha t Φ ' (z) = φ ( z ) and don’t forget t h e chain rule (c) Find P (−1 ≤ X ≤ 1) (d)Recall th at t h e probability th at Z is within one standard deviation of its mean is approximately 68%. W h at is t h e probability t ha t X is within one standard deviation of its mean?

14.Comp ute t h e median for t h e exponential distribution with parameter exponential distribution is defined on p a g e 62 of t h e t ext. )

λ. ( T h e

15. Let X and Y b e two random variables and let r, s, t, and u b e real numbers. (a) Show th at C ov( X + s, Y + u ) = Cov(X, Y ). (b) Show th at Cov(rX, tY ) = rtCov(X, Y ). (c) Show t ha t Co v( r X + s, tY + u ) = rtCov(X, Y ).

16. (Another Arithmetic Puzzle) Let X and Y b e two independent Bernoulli(.5) variables. Define S and T by: S = X + Y and T = X − Y.

random

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(a) Find t h e joint and marginal p m f ’ s for S and T . (b) Are S and T independent.

17. Tos s a fair coin 3 times. Let X = t h e number of h e a ds on t h e first toss, Y t h e total number of h e ad s on t h e last two t os se s, and F t h e number of h e a d s on t h e first two to s se s. (c) Give t h e joint probability table for X and Y . Compute Cov(X, Y ). (d) Give t h e joint probability table for X and F . Comp ute Cov(X, F ).

18. (M ore Central Limit Theorem ) T h e average IQ in a population is 100 with standard deviation 15 ( b y definition, IQ is normalized so this is t h e c as e ) . W hat is t h e probability th at a randomly se le c ted group of 100 p e op l e h a s an average IQ above 115?

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Solutions 1.

We build a full-house in s t a g e s and t h e number of ways to make e ach . count Σ 13 stage: S t ag e 1. C h oos e t h e rank of t h e .

1 . Σ pair: 4 S ta g e 2. Ch oo se t h e pair from th at rank, i.e. pick 2 of 4 .

2 . Σ cards: 12 S ta g e 3. Ch oo se t h e rank of t h e triple (from t h e remaining 12 .

1 . Σ ranks): 4 S ta g e 4. Ch oo se t h e triple from th at .

3 . Σ . Σ . Σ . Σ rank: 13 4 12 4 Number of ways to g e t a full1 2 1 3 . Σ house: 52 Number of ways to pick any 5 cards out of 5 . Σ. Σ. Σ. Σ 52: 13 4 12 14 1 2 . 1Σ 3 Probability of a full house : ≈ 0.00144 52 5 . 20Σ . 17Σ 2. (a) T he re are ways to c h o os ways to 3 2 . 15eΣ t h e 3 p e o pl e to s e t t h e table, c h o os e t h e 2 p e op l e to boil ways to c h o os e t h e p e o p l e to make sc one s. thewater, n 6 tand h e total number of ways to c h o os e p e op l eSofor t h e s e Σ. Σ tas ks is . 20Σ . 17 20!

15 20! 17! 15! = · · = = 775975200.

3 2 6 3! 17!

2! 15!

6!

3! 2! 6!

9! . 20Σ 9! (b) T h e number of ways to c h o os e 10 of t h e 20 p e opl e is T h e number of ways to 10 14 c ho o s e 10 p e op l e from t h e 14 Republicans .is Σ10 . So t h e probability th at you only c h o os e 10 Republicans is . Σ 14 14! . 10 Σ = 10! 4! ≈ 0.00542 20 10

20!

10! 10!

Alternatively, you could c h o os e t h e 10 p e o pl e in s e q u e nc e and say th at the re is a 14/20 probability th at t h e fi rst p er son is a Republican, the n a 13/19 probability th at t h e se c ond one is, a 12/18 probability th at third one is, etc. Thi s gives a probability of 14 13 12 11 10 9 8 7 6 5 · · · · Σ · 15 · 14 · 13 · 12 · 11 .

20 19 18 17 .16 (c) You can c ho o s e 1 Democrat in =61 6 ways, and you can c h o os e 9 Republicans in ways,can s o check t h e probability (You th at this is t h e s am e as t h e other answer given above.) . Σ 14! equals 6 · 14 6 · 9!

5! = 6 · 14! 10! . 20Σ9 = .

20!

10!9! 5!

10! 10!

10 20!

. 14Σ 9

Live Exam Helper Exam 1 Practice II, Spring 2014 3.

D is t h e disjoint union of D ∩ C and D ∩ C c .

C

Cc )

So, P ( D ∩ C ) + P ( D ∩ = P (D) ⇒ P ( D ∩ C c ) = P ( D ) − P ( D ∩ C ) = 0.45 − 0.1 = (We never u s e P ( C ) = 0.25.)

D D ∩ Cc

0.35.

0.1 0.45−0.1

We have t h e conditional 4.

Le t H i b e t h e event th at t h e i t h hand h a s one king. probabilities . Σ. Σ . Σ. Σ . Σ. Σ 2 24 4 48 3 36 1 12 1 12 1. 12 Σ ; P (H 3 |H 1 ∩ H 2 ) = P ( H 1 ) = . Σ52 ; P (H 2 |H 1 ) = . Σ 39 26 13 13 13 P (H 4 |H 1 ∩ H 2 ∩ H 3 ) = 1 P (H 1 ∩ H 2 ∩ H 3 ∩ H 4 ) = P (H .4 |H 1Σ∩. HΣ2 .∩ Σ H .3 ) PΣ (H . 3Σ|H . 1∩ Σ H 2 ) P (H 2 |H 1 ) P ( H 1 ) 2 24 3 36 4 48 1 12. 1Σ . 12 Σ . 1Σ 12 = .

26 39 52 13 13 13 5.

T h e following tree shows t h e setting. Stay 1 me ans t h e contestant was allowed to stay during t h e fi rst e p i s od e and stay 2 me ans t h e t h e y were allowed to stay during t h e second.

1 1

0 1/4

3/4

St ay 1 Leave 1 0 Bribe

Stay 2

Leave 2

Stay 1

1/3

Stay 2

1/3 2/3 Honest

2/3

Leave 1

Leave 2

Let’s name t h e relevant events: B = t h e contestant is bribing t h e j u dg e s H = t h e contestant is hones t (not bribing t h e j u d g e s ) S 1 = t h e contestant was allowed to stay during t h e fi rst e p i s od e S 2 = t h e contestant was allowed to stay during t h e se cond e p i s od e L 1 = t h e contestant was aske d to leave during t h e fi rst e p i s od e L 2 = t h e contestant was asked to leave during t h e se c ond 1 1 3 1 P ( S 1 ) = P (S 1 |B)P ( B ) + P (S 1 |H)P ( H ) = 1 · + · = .

4 3 4 2 e p i s od e P (B) 1/ 4 1 = .

We therefore have ( b y Bayes’ rule) P (B|S 1 ) = P (S 1 |B) P ( S )1 = 1 · 1/ 2 2 (a) We fi rst c om p ut e P ( S 1 ) using t h e law of total probability.

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(b) Using t h e tree we have t h e total probability of S 2 is 1 3 1 1 1 P (S2 ) = + · · = 4 4 3 3 3 P ( L 2∩ S ) 1 (c) We want to c o m p u te P ( L |S2 ) 1 = .

P (S1 ) From t h e calculation we did in part ( a) , P ( S 1 ) = 1/2. For t h e numerator, we have ( s e e t h e tre e ) 1 2 3 1 P (L 2 ∩ S 1 ) = P (L 2 ∩ S 1 |B)P ( B ) + P (L 2 ∩ S 1 |H)P ( H ) = 0 · + · = 1/6 Therefore P (L 2 |S 1 ) =

3

1 1/2

=

3

9

4

6

.

6.

You should write this out in a tree! (For example, s e e t h e solution to t h e next problem.)

We c o m p u t e all t h e p i e c e s n ee d ed to ap pl y Baye s’ rule. We’re given P (T |D) = 0.9 ⇒ P (T c |D) = 0.1, P (T |D c ) = 0.01 ⇒ P (T c |D c ) = 0.99. P ( D ) = 0.0005 ⇒ P ( D c ) = 1 − P ( D ) = 0.9995.

We u s e t h e law of total probability to c om p ut e P × (T 0.0005 ): P (D|T ) = P ( T | D) P ( D ) 0.9 = = 0.043 0.010445 c (T ) c P P (T ) = P (T |D) P ( D ) + P (T |D ) P ( D ) = 0.9 · 0.0005 + 0.01 · 0.9995 = P (T c |D) P ( D ) 0.1 × 0.0005 c = = 5.0 × 10 − 5 0.010445 P (D|T ) = P (Tc ) 0.989555 Now we u se Bayes’ rule to answer t h e questions: 7. B y tcan h e mutual i ndependence we have P ( A ∩ B ∩ C ) = P ( A ) P ( B ) P ( C ) = 0.06 P ( A ∩ C ) = P ( A ) P ( C ) = 0.15

P ( A ∩ B ) = P ( A ) P ( B ) = 0.12 P ( B ∩ C ) = P ( B ) P ( C ) = 0.2

We show this in t h e following Venn diagram A 0.09 0.09

0.060.14 0.06 0.14 0.21 C

B

Note that, for instance, P ( A ∩ B ) is split into two pieces. One of t h e p i e c e s is P ( A ∩ B ∩ which we know and t h e othe r we c o m p u t e as P ( A ∩ B ) −P ( A ∩ B ∩ C ) = 0.12 − 0.06 = 0.06.

C) T h e othe r intersections are similar.

We can read off t h e asked for probabilities from t h e diagram. (i ) P ( A ∩ B ∩ C c ) = 0.06 (ii) P ( A c∩ B c ∩ C ) = 0.09 (iii) P (A ∩ B ∩ C ) = 0.14.

8.

E ( X 2 ) = 27.

Use Va r ( X ) = E ( X 2 ) − E ( X ) 2 ⇒ 2 = E ( X 2 ) − 25 ⇒

1 .

It is e a s y to s e e t ha t (e .g. look at t h e probability tre e ) P ( 2 k ) = 2 k+1 ∞ Σ Σ 1 = ∞ . Technically, E ( X ) is undefi ned in this case . k 1 (b) E ( X ) = 2 = k =0 2 k +1 (c) Technically, E ( X ) 2is undefi ned in this case. B u t t h e value of ∞ tells u s w hat is wrong with t h e sch em e . Since th e average last b e t is infinite, I nee d to have an infinite amount of money in reserve.

9.

(a)

Thi s problem and solution is often referred to as t h e St. Petersburg paradox 10. (a) We have cdf of X ,

FX (x) =

Now for y ≥ 0, we have (b)

∫

x

λe 0

− λx dx = 1 − e .− λ x

√

2

− λ y,y we have Diff erentiating F Y ( y ) with re s p e c t to FY ( y ) = P (Y ≤ y ) = P ( X ≤ y )λ = −P1( −Xλ ≤y√ f Y ( y) = y 2 e.

2

11.

√

y) = 1 − e .

(a) Note: Y = 1 when X = 1 or X = −1, s o P (Y = 1) = P ( X = 1) + P ( X = −1).

Values y of Y 0 1 p mf pY ( y) 3/15 6/15 6/15

4

(b) and (c) To distinguish t h e distribution functions we’ll write F x and F Y .

Using t h e tables in part ( a ) and t h e defi nition F X ( a ) = P ( X ≤ a) 12. we g e t etc. a FX (a) FY ( a )

-1.5 1/15 0

3/4 6/15 3/15

7/8 6/15 3/15

1 10/15 9/15

1.5 10/15 9/15

5 1 1

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(i ) yes, discrete, (ii) no, (iii) no, (iv) no, ( v ) yes, continuous (vi) no (vii) yes, continuous, (viii) yes, continuous.

13. Normal Distribution: (a) We have F X( x ) = P ( X ≤ x ) = P (3 Z + 1 ≤ x ) = P ( Z ≤ (b)

x −1 x −1 ) = Φ . .

3 3 Σ

Diff erentiating with re s pe c t to x, we have

1

Since φ ( x ) = ( 2 π ) − 2 e −

d 1 f X( x ) = FX ( x ) = φ dx 2 3 x 2 , we conclude

x −1 . 3

Σ .

2 1 − (x − 1) e 2·32, f ( x ) = √3 2π which is t h e probability density function of t h e N (1, 9) distribution. Note: T h e arguments in ( a ) and ( b ) give a proof th at 3Z + 1 is a normal random variable with mean 1 and variance 9. S e e Problem S e t 3, Question 5. . Σ . Σ 2 2 (c) We have P (−1 ≤ X ≤ 1) = P − ≤ Z ≤ 0 = Φ(0) − Φ − ≈ 0.2475

X

3

(d)

3

Since E ( X ) = 1, Va r ( X ) = 9, we want P (−2 ≤ X ≤ 4). We have

P (−2 X ≤distribution 4) = P (−3 ≤ Z know ≤ 1) ≈(or 0.68. −λx. ≤ 14.T h e density for≤this is3Z f (≤ x )3)= =λ Pe (−1 We can c o m p u t e ) t ha t t h e − λ a distribution function is F ( a) = 1 − e . T h e median is t h e value of a su c h th at − λ a F ( a ) = .5. Thus, 1 − e = 0.5 ⇒ 0.5 = e − λ a ⇒ log(0.5) = −λa ⇒ a = l og ( 2 ) / λ .

15. (a) First note b y linearity of expectati on we have E ( X + s ) = E ( X ) + s, th us X + s − E ( X + s) = X − E ( X ) .

Likewise Y + u − E ( Y + u) = Y − E ( Y ). Now using t h e defi nition of covariance we g e t C o v( X + s, Y + u ) = E ( ( X + s − E ( X + s ) ) · (Y + u − E ( Y + u ) ) ) = E ( ( X − E ( X ) ) · (Y − E ( Y ) ) ) = Cov(X, Y ). (b) Thi s is very similar to part ( a ) .

Live Exam Helper We know E ( r X ) = r E ( X ) , s o r X − E ( r X ) = r ( X − E ( X ) ) . Likewise tY − E ( t Y ) = s(Y − E ( Y )) . Once again using t h e defi nition of covariance we g e t Cov(rX, tY ) = E ( ( r X − E ( r X ) ) ( t Y − E ( t Y ) ) ) = E(rt (X − E ( X ) ) ( Y − E(Y ) ) ) (Now we u s e linearity of expe ctation to pull out t h e factor of r t ) = rtE ((X − E ( X ) ( Y − E(Y )) ) ) = rtCov(X, Y ) (c) Th is is more of t h e same. We give t h e argument with far fewer algebraic details C ov( r X + s, tY + u ) = Cov(rX, tY ) ( b y part ( a ) ) = rtCov(X, Y ) ( b y part ( b ) ) 16. (Another Arithmetic Puzzle) (a) S = X + Y take s values 0, 1, 2 and T = X − Y takes values -1, 0, 1.

First we make two tables: t h e joint probability table for X and Y and a table given t h e values (S, T ) corresponding to values of ( X , Y ), e.g. ( X , Y ) = (1, 1) corresponds to (S, T ) = (2, 0).

Y

0

1

0 1

1/4 1/4

1/4 1/4

X\

X

\Y

0 1

0

1

0,0 1,1

0,-1 2,0

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Joint probabilities of X and Y Values of (S, T ) corresponding to X and Y We can u s e t h e two tables above to write t h e joint probability table for S and T . T h e marginal probabilities are given in t h e -1table.0 1 T

S\

0 1 2

1/4 0 0

1/4 0 0

0 1/4 1/4

1/2 1/4 1/4 1/4 1/4 1/2 1 Joint and marginal probabilities of S and T (b) No probabilities in t h e table are t h e produc t of t h e corresponding marginal probabilities. (T h i s is e asiest to s e e for t h e 0 entries.) So, S and T are not independent 17. (a) X and Y are independent, s o t h e table is c o m p u te d from t h e produ ct of t h e known marginal probabilities. Since t h e y are independent, Cov(X, Y ) = 0. (b) T h e samp le s p a c e is Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Y

\

X

0 1 2 PX

0

1

PY

1/8 1/4 1/8 1/2

1/8 1/4 1/8 1/2

1/4 1/2 1/4 1

Live Exam Helper P (X P (X P (X P (X P (X P (X

= = = = = =

0, F 0, F 0, F 1, F 1, F 1, F

= = = = = =

0) 1) 2) 0) 1) 2)

= = = = = =

P ({TTH, TTT } ) = 1/4.

P ({THH, THT } ) = 1/4.

0.

0.

P ({HTH, HTT } ) = 1/4.

P ({HHH, HHT } ) = 1/4.

Cov(X, F ) = E ( X F ) − E ( X ) E ( F ).

E ( X ) = 1/2, E ( F ) = 1, E ( X F ) =

F

\

X

0 1 2 PX

0

1

PF

1/4 1/4 0 1/2

0 1/4 1/4 1/2

1/4 1/2 1/4 1

Σ x y p(xi ,j y ) i = j 3/4.

⇒ Cov(X, F ) = 3/4 − 1/2 = 1/4.

18. (More Central Limit Theorem) Let X j b e t h e IQ of a randomly se l ec ted person. We are given E ( X j ) = 100 and σ X j = 15.

Let X b e t h e average of t h e IQ’s of 100 randomly se l e cte d people. Th e n we know Th is is eff ectively 0.

√

E ( X ) = 100and σ X = 15/ 100 = 1.5.

T h e proble m ask s for P ( X > 115). Standardizing we g e t P ( X > 115) ≈ P ( Z > 10).