B = {T3}, so P(AB)=1/12. ThereforeP(move at least 5 fields | you rolled a 3) = P(A|B) = P(A
B)/P(B) = (1/12)/(1/6) = 1/2Problem 1:
You are playing a board game. When it is your turn you flip a fair coin and then roll a fair die. If "heads" comes up you move as many fields as the die shows. If "tails" comes up you move twice as many fields as the die shows.
1) Write down the sample space
2) What is the probability of moving at least 5 fields?
3) What is the probability of moving at least 5 fields given you rolled a "3"?
4) What is the probability you rolled a "3" given that you moved at least 5 fields?
Problem 2:
We randomly select a person from WRInc. Let the rv X be the persons "satisfaction rating". Find the pmf of X
Problem 3:
Say we pick cards from a standard deck, one at a time.
1) What is the probability that the first card is a "diamond"?
2) What is the probability that the first card is a "diamond" and the second card is a "spades"?
3) What is the probability that the first three cards are all "diamonds"?
Problem 4:
We flip a fair coin repeatedly.
1) What is the probability of three "heads" in a row?
2) What is the probability that the first flip was heads given that two out of the first three flips were "heads"?
Problem 5:
A man comes home. He has 5 keys on his keychain, but he is so drunk he can't remember which is the key to the front door. So he randomly chooses one key after the other until he picks the right one. In fact he is so drunk he is just likely to pick the same key again even though he just tried it.
a) What is the probability that that he picks the right key on the first try?
b) What is the probability that that the first 5 tries are all wrong?
c) What is the mean and the standard deviation of the number of keys he tries until he picks the right one?
Problem 6
We let a computer randomly dial telephone numbers. When a person answers we make a note whether the person is male or female. If we talk to 70 people, what is the probability that between 32 and 37 are female? What are the mean and the standed deviation of the number of females in our sample? (you can assume that in this population there about as many males as females)
Problem 7 The number of words per page in the books of a certain author has a normal distribution with mean 237 and a standard deviation of 31.
1) If a page is selected at random from one of her books, what is the probability it will have between 200 and 250 words?
2) If a page is selected at random from one of her books, what is 90th percentile of the number of words?
3) If 20 pages are selected at random from her books, what is the probability the mean number of words per page will be more than 250?
4) If 20 pages are selected at random from her books, what is the first population quartile of the mean number of words?
Problem 1:
1) denote an outcome by T3 for "tails on the flip and a 3 on the roll". Then
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
so n(S)=12
2) Let A = "Move at least 5 fields" = {H5, H6, T3, T4, T5, T6}
so n(A)=6, therefore P(move at least 5 fields) = P(A) = 6/12 = 1/2
3) Let B = "rolled a 3" = {H3,T3}, so P(B) = 2/12 = 1/6, and AB = {T3}, so P(AB)=1/12. Therefore
P(move at least 5 fields | you rolled a 3) = P(A|B) = P(AB)/P(B) = (1/12)/(1/6) = 1/2
4) P(you rolled a 3 | you move at least 5 fields) = P(B|A) = P(BA)/P(A) = (1/12)/(1/2) = 1/6
Problem 2:
The rating goes from 1 to 5, so X takes values 1,2..,5
using Stat>Tables>Tally Individuell Variables, Satisfaction we find the frequencies, and so
Problem 3:
1) There are 13 diamonds in the 52 card deck, so P("diamond") = 13/52 = 1/4
2) P(first card is a "diamond" and the second card is a "spades") = P(first card is a "diamond") * P(the second card is a "spades" | first card is a "diamond") = 13/52 * 13/51
3) Let Di = "ith card is a diamond". Then
P(first three cards are all "diamonds") = P(D1D2D3) = P(D1)P(D2|D1)P(D3|D1D2) = 13/52*12/51*11/50
Problem 4:
1) Let Hi = "ith flip is "heads"". Then
P(three "heads" in a row) = P(H1H2H3) = P(H1)*P(H2)*P(H3) = 1/2*1/2*1/2 =1/8
Notice the difference to the problem above: here the flips (and therefore the events Hi) are independent whereas the draws froma deck are not, because we don't return the cards to the deck after we picked them.
2)
Problem 5:
If the rv X is the number of tries needed until he finds the right key, we have here X~G(1/5). so
a)P(first key is the right one) = P(X=1) = (1/5)·(4/5)1-1 = 1/5
b) P(first 5 tries are all wrong) = P(X>5) = 1-P(X≤5) =
1-((1/5)·(4/5)1-1+(1/5)·(4/5)2-1+(1/5)·(4/5)3-1+(1/5)·(4/5)4-1+(1/5)·(4/5)5-1)
=
1-(1/5)·(1+4/5+(4/5)2+(4/5)3+(4/5)4) =
1-(1/5)·(1+0.8+0.6400+0.5120+0.4096) = 1-(1/5)·3.362 = 0.3277
c) m=1/p=1/(1/5)=5
s=√(q)/p=√(4/5)/(1/5)=4.47
Problem 6
Let X be the number of females in the sample, then X~Bin(0.7,0.5) and so
P(32≤X≤37) = P(X≤37) - P(X≤31) = 0.5233
Also m = np = 70×0.5 = 35 and s = √npq = √70×0.5×0.5 = 4.18
Problem 7
1) We have X~N(237,31) and we want P(200<X<250), so
• (MINITAB) P(200<X<250) = P(X<250)-P(X<200) = 0.5462
• (tables) P(200<X<250) =
P((200-237)/31<(X-m)/s<(250-237)/31) =
P(-1.19<Z<0.42) =
P(Z<0.42) - P(Z<-1.19) =
0.6628 - 0.1170 = 0.5458
2) We want x such 0.9=P(X<x). So
• (MINITAB) x = 276.7
• (tables) 0.9=P(Z<z), so z=1.28 and x=m+sz = 237+31×1.28 = 276.7
3) We have 
~N(m, s) = N(237, 31/√20) = N(237, 6.93), so
• (MINITAB) P(X>250) = 1-P(X<250) = 0.030
• (tables) P(X>250) =
P((X-m)/s>(250-237)/6.93) =
P(Z>1.88) = 1- P(Z<1.88) = 1-0.9699 = 0.030
4) We want x such 0.25=P(<x), so
• (MINITAB) x= 232.3
• (tables) the number closest to 0.25, which is z=-0.67. So
x=m+sz = 237+6.93*(-0.67) = 232.6