51 terms

if the experiment is flipping a coin, the sample space is

S = {H,T}

(Packet 1)

(Packet 1)

if we flip two coins the sample space is

S={(H,H) (H,T) (T,H) (T,T)}

(Packet 1)

(Packet 1)

pick a die from a black box containing three dices: red, green, and yellow. the sample space is

S={R,G,Y}

(Packet 1)

(Packet 1)

if you roll a dice, what is the sample space

S = {1,2,3,4,5,6}

(Packet 1)

(Packet 1)

toss a coin until you see a head. sample space

S = {H, TH, TTH, TTTH...} note that S has infinite many outcomes

(Packet 1)

(Packet 1)

when flipping two coins, what is the event that the first coin is heads

S = {(H,H) (H,T) (T,H) (T,T)}

Event = {(H,H) (H,T)}

(Packet 1)

Event = {(H,H) (H,T)}

(Packet 1)

roll a dice. S ={1,2,3,4,5,6}

let A = {1,3,5}

let B = {1,2,3}

find: A^c

b^c

ANB (a and b)

A U B

(ANB)^c

(AUB)^c

let A = {1,3,5}

let B = {1,2,3}

find: A^c

b^c

ANB (a and b)

A U B

(ANB)^c

(AUB)^c

A^c = {2,4,6}

B^c ={4,5,6}

ANB (a and b) = {1,3}

A U B = {1,2,3,5}

(ANB)^c = {2,4, 5,6}

(AUB)^c = {4,6}

(Packet 1)

B^c ={4,5,6}

ANB (a and b) = {1,3}

A U B = {1,2,3,5}

(ANB)^c = {2,4, 5,6}

(AUB)^c = {4,6}

(Packet 1)

roll a dice, what is the probability of 4

S = {1,2,3,4,5,6}

so p = 1/6

(Packet 1)

so p = 1/6

(Packet 1)

roll two dice, what is the event that dice 1 + dice 2 = 5 (what is P(A)?)

A ={(1,4) (4,1) (2,3) (3,2)}

so P(A) = 1/36 + 1/36 + 1/36 + 1/36 = 1/9

(Packet 1)

so P(A) = 1/36 + 1/36 + 1/36 + 1/36 = 1/9

(Packet 1)

a card is delt from a well shuffled deck. what is the change of getting a Jack?

1/52 = probability of getting one card

4/52 = probability of getting a Jack

(Packet 1)

4/52 = probability of getting a Jack

(Packet 1)

let P(A) = 1/3 and P(B) = 1/2

Determine the value of P(BA^c) for the following conditions:

1. A and B are disjoint

2. A is contained in B

3. P(AB) = 1/8

Determine the value of P(BA^c) for the following conditions:

1. A and B are disjoint

2. A is contained in B

3. P(AB) = 1/8

1. if A and B are disjoint then B is contained in A^c. so BA^c = B N A^c (B and A^c) which means P(BA^c) = P(B) = 1/2

2. if A is contained in B then B = (BA)U(BA^c) = A U BA^c and A N BA^c = empty set so P(B)= P(A) + P(BA^c) which means P(BA^c) = P(B)-P(A) = 1/2 -1/3 = 1/6

3. note B = (BA)U(BA^c) = A U BA^c and A N BA^c = empty set

so P(B) = P(BA) + P(BA^c) which means P(BA^c) = P(B) - P(BA) = 1/2 - 1/8 = 3/8

(Packet 1)

2. if A is contained in B then B = (BA)U(BA^c) = A U BA^c and A N BA^c = empty set so P(B)= P(A) + P(BA^c) which means P(BA^c) = P(B)-P(A) = 1/2 -1/3 = 1/6

3. note B = (BA)U(BA^c) = A U BA^c and A N BA^c = empty set

so P(B) = P(BA) + P(BA^c) which means P(BA^c) = P(B) - P(BA) = 1/2 - 1/8 = 3/8

(Packet 1)

a patient with a sore throat and low grade fever is believed to have either bacterial infection or viral infection. The probability of bacterial infection is 0.7 and viral is 0.4. What is the probability that the patient has both bacterial and viral infection?

BI = bacterial infection

VI = Viral infection

P(BI) = 0.7 and P(VI) = 0.4

the statement "The patient is believed to have either bacterial or viral infection" means P(BI U VI) = 1

P(BI U VI) = P(BI) + P(VI) - P(BI N VI) so

P(BI N VI) = 0.7 + 0.4 -1 = 0.1

(Packet 1)

VI = Viral infection

P(BI) = 0.7 and P(VI) = 0.4

the statement "The patient is believed to have either bacterial or viral infection" means P(BI U VI) = 1

P(BI U VI) = P(BI) + P(VI) - P(BI N VI) so

P(BI N VI) = 0.7 + 0.4 -1 = 0.1

(Packet 1)

a student applies to X and Y schools. x = 0.7 and Y = 0.4. he suspects that there is a .75 chance that at least one school will reject hi. what is the probability that at least one will accept him

P(AX) = 0.7

P(AY) = 0.4

the event that one will accept him = P(AX U AY)

we use P(AX U AY) = P(AX) + P(AY) - P(AX N AY)

we know P(AX N AY) is he is accepted into both schools so P(AX N AY)^c = 0.75

so P(AX N AY) = 1 -0.75 = .25

so 0.7 + 0.4 -0.25 = 0.85

(Packet 1)

P(AY) = 0.4

the event that one will accept him = P(AX U AY)

we use P(AX U AY) = P(AX) + P(AY) - P(AX N AY)

we know P(AX N AY) is he is accepted into both schools so P(AX N AY)^c = 0.75

so P(AX N AY) = 1 -0.75 = .25

so 0.7 + 0.4 -0.25 = 0.85

(Packet 1)

2 balls are randomly chosen from a bowl containing 6 white and 5 black balls. what is the probability that one of the drawn balls is white and the other black?

11 choices for the first ball

10 choices for the second

so a total of 11*10 = 110 choices

for a white ball and then black ball that is 6*5 = 30

for a black ball then a white ball that is 5*6 = 30

so there is a total of 30+30 = 60 outcomes

so it is 60/110 or 6/11

(Packet 2)

10 choices for the second

so a total of 11*10 = 110 choices

for a white ball and then black ball that is 6*5 = 30

for a black ball then a white ball that is 5*6 = 30

so there is a total of 30+30 = 60 outcomes

so it is 60/110 or 6/11

(Packet 2)

throw 2 dice, red and green, we want an (X,Y) outcome. how many pairs will we have

1. throw red dice

2. throw green dice

6*6= 36

(Packet 2)

2. throw green dice

6*6= 36

(Packet 2)

how many distinct arrangements of the letters A, B, and C are possible

1. put A into a spot (3)

2. put B into a spot (2)

3. put C into a spot (1)

3**2**1 = 6

(Packet 2)

2. put B into a spot (2)

3. put C into a spot (1)

3

(Packet 2)

a committee of chair, secretary and treasurer is selected from 20 people, how many committees are possible

1. select chair (20)

2. select secretary (19)

3. select treasurer (18)

20**19**18 = 6840 combinations

(Packet 2)

2. select secretary (19)

3. select treasurer (18)

20

(Packet 2)

there are 62 students in a classroom. what is the probability that two have the same birthday

there are 62 students

there are 356 days in a yar

(1 -1/365)(1-2/365)...(1-(62-1)/365) = 0.005

so P(A) = 1 -P(A^c) so we get 0.995

(Packet 2)

there are 356 days in a yar

(1 -1/365)(1-2/365)...(1-(62-1)/365) = 0.005

so P(A) = 1 -P(A^c) so we get 0.995

(Packet 2)

Suppose that a committee composed of 8 people is to be selected from a group of 20. what is the number of different groups

20!/8!(20-8)! = 20!/8!12! = 125970

(Packet 2)

(Packet 2)

how many committees with 2 republicans, 2 democrats and 3 independents can be formed from a group of 6 republicans, 5 democrats, and 4 independents

1. select 2 republicans (6!/2!4!)

2. select 2 democrats (5!/2!3!)

3. select 3 independents (4!/3!1!)

those multiplied together gets you 600

(Packet 2)

2. select 2 democrats (5!/2!3!)

3. select 3 independents (4!/3!1!)

those multiplied together gets you 600

(Packet 2)

suppose a coin is tossed 10 times. what is the probability of:

1. obtaining exactly 3 heads

2. obtaining 3 or fewer heads

1. obtaining exactly 3 heads

2. obtaining 3 or fewer heads

the total outcomes of tossing a coin 10 times is 2^10

1. we want 3 heads so we want 10!/3!7!

so (10!/7!3!)/2^10 = .1172

2. b is the event we have 3 or fewer heads

x is the event of exactly 2 heads (10!/2!8!)

y is the event of exactly 1 head (10!/1!9!)

z is the event of exactly 0 heads (10!/0!10!)

so (10!/7!3!)/2^10 + (10!/2!8!)/2^10 + (10!/1!9!)/2^10 + (10!/0!10!)/2^10 = 0.1719

(Packet 2)

1. we want 3 heads so we want 10!/3!7!

so (10!/7!3!)/2^10 = .1172

2. b is the event we have 3 or fewer heads

x is the event of exactly 2 heads (10!/2!8!)

y is the event of exactly 1 head (10!/1!9!)

z is the event of exactly 0 heads (10!/0!10!)

so (10!/7!3!)/2^10 + (10!/2!8!)/2^10 + (10!/1!9!)/2^10 + (10!/0!10!)/2^10 = 0.1719

(Packet 2)

we want to divide 12 bridge players into 3 rooms to play the game. what is the number possible of arrangements

1. fill room 1 (12!/4!8!)

2. fill room 2 (8!/4!4!)

3. fill room, 3 (1)

(12!/4!8!)**(8!/4!4!)**1 = 34650

or:

12!/4!4!4! = 34650

(Packet 2)

2. fill room 2 (8!/4!4!)

3. fill room, 3 (1)

(12!/4!8!)

or:

12!/4!4!4! = 34650

(Packet 2)

how many distinct arrangements can be formed from the letters "Mississippi"

we have 11 letters. each arrangement is a partition, we need to find 1 position for m, 4 for I, 4 for s, and 2 for p

you get 11!/1!4!4!2! = 34650

(Packet 2)

you get 11!/1!4!4!2! = 34650

(Packet 2)

what is the coefficient of x^2y^3 in the expansion of (1 + x + y)^100

a = 1

b = 2

c = 3 and n = 100

so 100-2-3=95

then 100!/95!3!2! = 485100

(Packet 2)

b = 2

c = 3 and n = 100

so 100-2-3=95

then 100!/95!3!2! = 485100

(Packet 2)

roll a dice. let A denote the event that we get 1 and B denote the event that we get an odd number. Assume we get an odd number, what is the probability that it is 1?

b = {1,3,5}

if B occurs, then 1,3,5 all have the same amount of chances of occurring, so the probability that it is 1 is 1/3

(Packet 3)

if B occurs, then 1,3,5 all have the same amount of chances of occurring, so the probability that it is 1 is 1/3

(Packet 3)

a game host has $100, $10, and $1 in three cases. the player selects one case, the host opens one of the two remaining cases. lets say it is not the $100. you are allowed to now switch your case. Do you?

A = the first chosen case has $100

before one case is opened, we know P(A) = 1/3

B = the second case does not contain $100 (the opened case)

so what is P(A | B)?

P(A^c) = 2/3

P(B|A) = 1

P(B|A^c) = 1/2

Using Bayes Theorem:

P(A|B) = (P(B|A)P(A)) / (P(B|A)P(A) + P(B |A^c)P(A^c)

= (1**(1/3))/(1**(1/3) + (1/2)*(2/3))

= 1/2

So you want to switch the cases

(Packet 3)

before one case is opened, we know P(A) = 1/3

B = the second case does not contain $100 (the opened case)

so what is P(A | B)?

P(A^c) = 2/3

P(B|A) = 1

P(B|A^c) = 1/2

Using Bayes Theorem:

P(A|B) = (P(B|A)P(A)) / (P(B|A)P(A) + P(B |A^c)P(A^c)

= (1

= 1/2

So you want to switch the cases

(Packet 3)

a box contains 10 white, 5 yellow, and 10 black balls. a ball is chosen at random, and it is not black. what is the probability that it is yellow?

A = selected marble is yellow

B = the selected marble is not black

we need P(A | B)

so P(B) = (10+5)/(10+5+10) = 0.6

we see that AB = A so

P(AB) = P(A) = 5/(10+5+10) = 0.2

so P(A|B) = 0.2/0.6 = 1/3

(Packet 3)

B = the selected marble is not black

we need P(A | B)

so P(B) = (10+5)/(10+5+10) = 0.6

we see that AB = A so

P(AB) = P(A) = 5/(10+5+10) = 0.2

so P(A|B) = 0.2/0.6 = 1/3

(Packet 3)

a group is throwing a party for people with at least one son. if you are invited and have 2 kids, what is the probability that they are both sons?

for having two children, there are 4 possible combinations (B,B)(B,G)(G,B)(G,G)

each has a 1/4 probability of occurring

A = event she has 2 sons, A = (B,B)

B = event that she was invited to the party, B = (B,B)(B,G)(G,B)

so P(A)= 1/4 and P(B)= 3/4

AB = A so that also = 1/4

then P(A|B) = (.25)/(.75) = 1/3

(Packet 3)

each has a 1/4 probability of occurring

A = event she has 2 sons, A = (B,B)

B = event that she was invited to the party, B = (B,B)(B,G)(G,B)

so P(A)= 1/4 and P(B)= 3/4

AB = A so that also = 1/4

then P(A|B) = (.25)/(.75) = 1/3

(Packet 3)

roll a dice, A = a prime number, B = an even number, what is P(A|B)?

A = {2,3,5}

B = {2,4,6}

AB = {2} (AB is the event that both A and B occur)

P(A|B) = P(AB)/P(B) = (1/6)/(3/6) = 1/3

P(A|B) does not equal P(A)/P(B) because .5/.5 = 1

(Packet 3)

B = {2,4,6}

AB = {2} (AB is the event that both A and B occur)

P(A|B) = P(AB)/P(B) = (1/6)/(3/6) = 1/3

P(A|B) does not equal P(A)/P(B) because .5/.5 = 1

(Packet 3)

a box has 7 black and 5 white balls. we randomly select 2 balls. what is the probability that both are black.

B = the first ball drawn is black

A = the second ball drawn is black

we want P(AB)

P(B) = 7/12

P(A |B) = 6/11 because there are 6 black balls left and 5 white

so P(AB) = 7/12 * 6/11 = 7/22

(Packet 3)

A = the second ball drawn is black

we want P(AB)

P(B) = 7/12

P(A |B) = 6/11 because there are 6 black balls left and 5 white

so P(AB) = 7/12 * 6/11 = 7/22

(Packet 3)

select 4 balls from a box one at a time. the box contains r red balls and b blue balls. what is the probability of obtaining the sequence of outcomes, red, blue, red blue?

Rj = the event that the jth ball is red (j= 1,2,3,4...)

Bj = the event that the jth ball is blue (j = 1,2,3,4...)

we need to compute P(R1B2R3B4)

= P(R1)P( B2|R1)P(R3|B2R1)P(B4|R3B2R1)

= (R/R+B)** (B/(R-1+B)) ** ((R-1)/(R-1+B-1)) * ((B-1)/ (R-2+B-1))

= (RB(R-1)(B-1))/((R+B)(R+B-1)(R+B-2)(R+B-3))

For R = 10 and B = 20,

(10**20**9**19)/(30**29**28**27) = 0.052

(Packet 3)

Bj = the event that the jth ball is blue (j = 1,2,3,4...)

we need to compute P(R1B2R3B4)

= P(R1)P( B2|R1)P(R3|B2R1)P(B4|R3B2R1)

= (R/R+B)

= (RB(R-1)(B-1))/((R+B)(R+B-1)(R+B-2)(R+B-3))

For R = 10 and B = 20,

(10

(Packet 3)

roll a dice A = {2,4,6}

B = {1,2,3,4}

are A and B independent?

B = {1,2,3,4}

are A and B independent?

Must show that:

P(A) = P(A|B)

P(B) = P(B|A)

P(ANB) (A and B) = P(A)P(B)

P(A) = 3/6 = 1/2

P(B) = 4/6 = 2/3

AB = {2,4}

so P(AB) = 1/3

so P(AB) = P(A)P(B)

1/3 = (1/2)(2/3)

By definition, A and B are independent

(Packet 3)

P(A) = P(A|B)

P(B) = P(B|A)

P(ANB) (A and B) = P(A)P(B)

P(A) = 3/6 = 1/2

P(B) = 4/6 = 2/3

AB = {2,4}

so P(AB) = 1/3

so P(AB) = P(A)P(B)

1/3 = (1/2)(2/3)

By definition, A and B are independent

(Packet 3)

roll two dice. A = the event that the sum of the dice is 6. B = the event that the first dice is 4. Are A and B independent?

A = {(4,2)(2,4)(5,1)(1,5)(3,3)}

P(A) = 5/36

B = {(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)}

P(B) = 6/36 = 1/6

AB = {(4,2)}

P(AB) = 1/36

1/36 ? (5/36)(1/6)

1/36 < (5/36)(1/6)

so A and B are dependent

(Packet 3)

P(A) = 5/36

B = {(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)}

P(B) = 6/36 = 1/6

AB = {(4,2)}

P(AB) = 1/36

1/36 ? (5/36)(1/6)

1/36 < (5/36)(1/6)

so A and B are dependent

(Packet 3)

a machine produces a defective item with probability p and produces a non defective item q = 1 - p. 6 items are selected at random and inspected. the outcomes for these 6 items are independent. compute the probability that exactly 2 of the 6 items are defective

let Dj = the event that the jth item is defective

Nj = denote the event that the jth item is non-defective

A = the event that there are 2 defective items and 4 non-defective

one possible outcome is: N1D2N3N4D5N6

by the rules of independence,

P(N1D2N3N4D5N6) = P(N1)P(D2)P(N3)P(N4)P(D5)P(N6)

= qpqqpq = p^2 * q^4

there are 6 choose 2 outcomes and each has the same probability. thus the desired probability is 6!/2!3!p^2q^4 = 15p^2q^4

(Packet 3)

Nj = denote the event that the jth item is non-defective

A = the event that there are 2 defective items and 4 non-defective

one possible outcome is: N1D2N3N4D5N6

by the rules of independence,

P(N1D2N3N4D5N6) = P(N1)P(D2)P(N3)P(N4)P(D5)P(N6)

= qpqqpq = p^2 * q^4

there are 6 choose 2 outcomes and each has the same probability. thus the desired probability is 6!/2!3!p^2q^4 = 15p^2q^4

(Packet 3)

a blood test is 95% effective in detecting a disease when it is present. the test also yields a false positive result for 1% of the healthy people tested. if 0.5 % of the population actually have the disease, what is the probability that a person has the disease when his test result is positive?

D = has the disease

E = positive test result

we want to find P(D|E)

P(E|D) = 0.95

P(E|D^c) = 0.01 (the false positive rate)

P(D) = 0.005 (the population that actually has the disease)

P(D^c) = 1 - P(D) = 0.995

using Bayes Theorem:

P(D|E) = (P(E|D)P(D))/(P(E|D)P(D) + P(E|D^c)P(D^c)

= .95**.005/(.95**.005 + .01*.995)

= 32% of the people whose test results are positive actually have the disease

(Packet 3)

E = positive test result

we want to find P(D|E)

P(E|D) = 0.95

P(E|D^c) = 0.01 (the false positive rate)

P(D) = 0.005 (the population that actually has the disease)

P(D^c) = 1 - P(D) = 0.995

using Bayes Theorem:

P(D|E) = (P(E|D)P(D))/(P(E|D)P(D) + P(E|D^c)P(D^c)

= .95

= 32% of the people whose test results are positive actually have the disease

(Packet 3)

a plane crashed and is presumed that it cashed in 2 regions. let 1 - Ai be the probability that the plane will be found if the plane is in region i, for i = 1,2. so Ai is the overlook probability. for instance A1 = 0.1 and A2 = 0.05. what is the probability that the plane is in region 2 given that a search in region 1 is unsuccessful?

E = the event that he search is unsuccessful in region 1

Ri denotes the event that the plane is in region i, for i = 1,2

we want to find P(R2|E)

using Bayes theorem:

P(R2|E) = (P(E|R2)P(R2))/(P(E|R1)P(R1) + P(E|R2)P(R2))

= (1**(1/2))/(Ai**(1/2) + 1*(1/2))

= 1/(1+Ai)

= 1/(1 + 0.1)

= 10/11

= 0.91

(Packet 3)

Ri denotes the event that the plane is in region i, for i = 1,2

we want to find P(R2|E)

using Bayes theorem:

P(R2|E) = (P(E|R2)P(R2))/(P(E|R1)P(R1) + P(E|R2)P(R2))

= (1

= 1/(1+Ai)

= 1/(1 + 0.1)

= 10/11

= 0.91

(Packet 3)

three cards are placed in a box. one is red on both sides, the second is blue on both and the third is red on one side and blue on the other. one card is selected at random and placed on the table. suppose the color that is showing is red. what is the probability that the color underneath is also red?

A = the event the bottom side of the selected card is red

B = the event that the top side of the selected card is red

we want P(A|B)

C1 = the selected card is red/red

C2 = the selected card is blue/blue

C3 = the selected card is red/blue

P(A|B) = P(AB)/P(B)

= P(C1)/ (P(B|C1)P(C1) + P(B|C2)P(C2) + P(B|C3)P(C3))

= (1/3)/( 1**(1/3) + 0**(1/3) + (1/2)*(1/3))

= 2/3

(Packet 3)

B = the event that the top side of the selected card is red

we want P(A|B)

C1 = the selected card is red/red

C2 = the selected card is blue/blue

C3 = the selected card is red/blue

P(A|B) = P(AB)/P(B)

= P(C1)/ (P(B|C1)P(C1) + P(B|C2)P(C2) + P(B|C3)P(C3))

= (1/3)/( 1

= 2/3

(Packet 3)

the 38 equally likely positions on a roulette wheel are numbered 0,00,1,2... 36. the 0 and 00 are green the rest half are black and half are red. find the probability that the outcome of a single spin is:

1. red

2. not black

3. numbers 25-36

1. red

2. not black

3. numbers 25-36

38 equal possibilities so Pi = 1/38

1. red P(A) = (1/38)*(18) = 9/19

2. not black P(A) = (1/38)* 20 = 10/19

3. P(A) = 1/38 *12 = 6/19

(homework 1)

1. red P(A) = (1/38)*(18) = 9/19

2. not black P(A) = (1/38)* 20 = 10/19

3. P(A) = 1/38 *12 = 6/19

(homework 1)

60 % of students of a certin school wear neither a ring nor a necklace. 20% wear and ring and 30% wear a necklace. if one student is chosen randomly, what is the probability that this student is wearing:

1. a ring or a necklace

2. a ring and a necklace

1. a ring or a necklace

2. a ring and a necklace

X = wear a ring = 0.2

Y = wear a necklace = 0.3

1. 0.2 + 0.3 - 0.1 = P(X U Y)

= 40% probability a student has a necklace or ring

2. P(X^c N Y^c) = 0.6 = wear neither

P(X) + P(Y) - P(X N Y) = P(XUY)

so P(XUY) = 1 - P(X^c N Y^c)

0.2 +0.3 - P(XNY) = 1 -0.6

=0.1 probability a student is wearing a necklace and ring

(Homework 1)

Y = wear a necklace = 0.3

1. 0.2 + 0.3 - 0.1 = P(X U Y)

= 40% probability a student has a necklace or ring

2. P(X^c N Y^c) = 0.6 = wear neither

P(X) + P(Y) - P(X N Y) = P(XUY)

so P(XUY) = 1 - P(X^c N Y^c)

0.2 +0.3 - P(XNY) = 1 -0.6

=0.1 probability a student is wearing a necklace and ring

(Homework 1)

if 50% of families in a city subscribe to the morning newspaper, 65% of families get the afternoon one, 85% get at least one of the two. what is the percentage of the families subscribe to both?1

P(X) = morning paper = 0.50

P(Y) = afternoon paper = 0.65

one of the two papers = 0.85 = P(XUY)

P(X) + P(Y) - P(XNY) = P(XUY)

.5 + .65 - P(XNY) = .85

30 % of families get both

(Homework 1)

P(Y) = afternoon paper = 0.65

one of the two papers = 0.85 = P(XUY)

P(X) + P(Y) - P(XNY) = P(XUY)

.5 + .65 - P(XNY) = .85

30 % of families get both

(Homework 1)

tom thinks 80% he will get an A in physics and with 60% he will get an A in chemistry. he also guesses that the chance he will get A in both is 0.5. what is the change he will get exactly one A?

X = A in physics = 0.8

Y = A in chemistry = 0.6

P(X N Y) = A in both = 0.5

P(X) + P(Y) - P(XNY) = P(XUY)

= 0.8 + 0.6 - 0.5

= 0.9

P(exactly 1 A) = P(XUY) - P(XNY)

0.9 - 0.5

= 0.4

40% chance of getting exactly 1 A

(Homework 1)

Y = A in chemistry = 0.6

P(X N Y) = A in both = 0.5

P(X) + P(Y) - P(XNY) = P(XUY)

= 0.8 + 0.6 - 0.5

= 0.9

P(exactly 1 A) = P(XUY) - P(XNY)

0.9 - 0.5

= 0.4

40% chance of getting exactly 1 A

(Homework 1)

if 2 balls are randomly selected from a box containing 20 red and 15 green balls, what is the probability that the chosen balls are the same color?

20 red

15 green

Part 1:

1. select first ball (35)

2. select second ball (34)

total number of combinations = (35*34)/2 = 595

Part 2:

1. pick a red ball first (20)

2. pick a red ball second (19)

total possibilities of getting (r,r) = (20*19)/2 = 190

part 3:

1. pick a green ball first (15)

2. pick a green ball second (14)

total possibilities of getting (g,g) = (15*14)/2 = 105

total possibilities of getting the same color:

(190+105)/595 = 50%

(Homework 2)

15 green

Part 1:

1. select first ball (35)

2. select second ball (34)

total number of combinations = (35*34)/2 = 595

Part 2:

1. pick a red ball first (20)

2. pick a red ball second (19)

total possibilities of getting (r,r) = (20*19)/2 = 190

part 3:

1. pick a green ball first (15)

2. pick a green ball second (14)

total possibilities of getting (g,g) = (15*14)/2 = 105

total possibilities of getting the same color:

(190+105)/595 = 50%

(Homework 2)

how many ways to arrange the letters in "s,s,s,t,t,t,i,i,a,c" if the arrangement is random , what is the probability of spelling the word "statistic"?

we have 10 letters to fill 10 positions. we can divide the 10 positions into 5 groups,

s = 3, t = 3, i = 2, a = 1, c = 1

number of total possibilities =

10!/ (3!3!2!1!1!)

probability of getting the word statistic = 1/(10!/(3!3!2!1!1!))

(Homework 2)

s = 3, t = 3, i = 2, a = 1, c = 1

number of total possibilities =

10!/ (3!3!2!1!1!)

probability of getting the word statistic = 1/(10!/(3!3!2!1!1!))

(Homework 2)

suppose that 9 red, 5 yellow, 8 blue and 12 black flags are to be hung in a row. how many different arrangements can be made?

total number of flags: 34

9 red

8 blue

5 yellow

12 black

number of possible arrangements:

34!/(9!5!8!12!)

(Homework 2)

9 red

8 blue

5 yellow

12 black

number of possible arrangements:

34!/(9!5!8!12!)

(Homework 2)

roll a dice 36 times. compute the probability of obtaining exactly 6 "6"

36 rolls on a dice

total possibilities: 6^(36)

to get exactly 6 "6": 36! / (30!6!)

for the left over possibilities: 5^(30)

so the probability of getting exactly 6 "6"

((5^30)*( 36!/30!6!) ) / (6^36)

= 18%

(Homework 2)

total possibilities: 6^(36)

to get exactly 6 "6": 36! / (30!6!)

for the left over possibilities: 5^(30)

so the probability of getting exactly 6 "6"

((5^30)*( 36!/30!6!) ) / (6^36)

= 18%

(Homework 2)

suppose you are asked to answer 9 out of 12 questions in an exam. you must answer at least 3 of the first 4 questions, in how many ways can you choose 9 questions to answer?

1. you can choose 3 from the first 4: (4!/3!1!)

then you can choose 6 from the last 8: (8!/6!2!)

total possibilities of step 1: (4!/3!1!)*(8!/6!2!)

2. you can choose 4 from the first 4: (4!/4!0!)

then you can choose 5 from the last 8: (8!/5!3!)

total possibilities of step 2: (4!/4!0!)(8!/5!3!)

the number of different ways to answer the 9 questions:

add those together so you get 168

(Homework 2)

then you can choose 6 from the last 8: (8!/6!2!)

total possibilities of step 1: (4!/3!1!)*(8!/6!2!)

2. you can choose 4 from the first 4: (4!/4!0!)

then you can choose 5 from the last 8: (8!/5!3!)

total possibilities of step 2: (4!/4!0!)(8!/5!3!)

the number of different ways to answer the 9 questions:

add those together so you get 168

(Homework 2)

a basket has 10 black and 15 white balls. we randomly pick 2 balls. using conditional arguments to compute the probability that we select 1 white ball and 1 black ball.

two parts:

1. first ball is white (15/25)

then second ball is black (10/24)

want P(AB) = (15/25)*(10/24) = .25

2. first ball is black (10/25)

then second ball is white (15/24)

want P(AB) = (10/25)*(15/24) = .25

.25 + .25 = 1/2

the probability of getting 1 white and 1 black is 50%

(Homework 3)

1. first ball is white (15/25)

then second ball is black (10/24)

want P(AB) = (15/25)*(10/24) = .25

2. first ball is black (10/25)

then second ball is white (15/24)

want P(AB) = (10/25)*(15/24) = .25

.25 + .25 = 1/2

the probability of getting 1 white and 1 black is 50%

(Homework 3)

roll 2 dice. let A = the sum is 7

B = the first dice is 4

are A and B independent?

B = the first dice is 4

are A and B independent?

A = the sum is 7

B = the first is 4

P(A) = 6/36 = 1/6

P(B) = 6/36 = 1/6

P(AB) = 1/36

P(A)P(B) = P(AB) then independent

(1/6)(1/6) = 1/36

they are independent

(Homework 3)

B = the first is 4

P(A) = 6/36 = 1/6

P(B) = 6/36 = 1/6

P(AB) = 1/36

P(A)P(B) = P(AB) then independent

(1/6)(1/6) = 1/36

they are independent

(Homework 3)

40 % of students in college are female and suppose further that 15% of the female and 10% of the male students are smokers. we randomly choose a smoker, what is the probability that the student is female?

A = the student is female

B = the student is a smoker

want P(A|B)

P(B|A) = .15

P(A) = 0.4

P(B|A^c) = .85

P(A^c) = 0.6

P(A|B) = .15**.4/ (.15**.4 + .85*.6)

= 11% Chance the student is female

(Homework 3)

(I think its right)

B = the student is a smoker

want P(A|B)

P(B|A) = .15

P(A) = 0.4

P(B|A^c) = .85

P(A^c) = 0.6

P(A|B) = .15

= 11% Chance the student is female

(Homework 3)

(I think its right)

we have 2 boxes, box A = 1 black and 2 white marbles

box B = 3 black and 2 white marbles

one box is selected at random and a marble is drawn from the selected box. suppose the marble selected is black, what is the chance that the chosen box is B?

box B = 3 black and 2 white marbles

one box is selected at random and a marble is drawn from the selected box. suppose the marble selected is black, what is the chance that the chosen box is B?

A= box is B

B = the marble chosen is black

want P(A|B)

P(A) = 1/2

P(B) = (1/2)*(1/3) + (1/2)(3/5) = 14/30

P(A^c) = 1/2

P(B|A) = 3/5

P(B|A^c) = 1/3

P(A|B) = ((3/5)(1/2)) / ( (3/5)(1/2) + (1/3)(1/2) )

= 64%, So if you select a black marble there is this chance its from box B

(Homework 3)

B = the marble chosen is black

want P(A|B)

P(A) = 1/2

P(B) = (1/2)*(1/3) + (1/2)(3/5) = 14/30

P(A^c) = 1/2

P(B|A) = 3/5

P(B|A^c) = 1/3

P(A|B) = ((3/5)(1/2)) / ( (3/5)(1/2) + (1/3)(1/2) )

= 64%, So if you select a black marble there is this chance its from box B

(Homework 3)

30% of bottles produces in a plant are defective. if a bottle is defective, the probability is .9 that it will be removed from the line, if a bottle is not defective, the probability is 0.2 that the inspector will think it is defective and remove it. if a bottle is removed from the line, what is the probability it is defective?

A = bottle is defective

B = inspector will remove it from the line

want P(A|B)

P(B|A) = .9

P(B|A^c) = .2

P(A) = .3

P(A^c) = .7

P(A|B) = .9**.3/(.9**.3 + .2*.7)

= 66% if the bottle is pulled it is defective

(Homework 3)

B = inspector will remove it from the line

want P(A|B)

P(B|A) = .9

P(B|A^c) = .2

P(A) = .3

P(A^c) = .7

P(A|B) = .9

= 66% if the bottle is pulled it is defective

(Homework 3)