Download Probability Problems III

Probability Problems III 1) Suppose you have a twelve‐sided die (each face has a value from 1 to 12) and a four‐sided die (each face has a value from 1 to 4) and you roll them together. We define Event A as the subtraction of the four‐sided die from the twelve sided die. 1 2 3 4 5 6 7 8 9 10 11 12
a) Create the sample space of Event A 1 0 1 2 3 4 5 6 7 8 9 10 11
b) What is P(A < 0)? 6/48 = 1/8 2 ‐1 0 1 2 3 4 5 6 7 8 9 10
c) What is the P(A = 0)? 4/48 = 1/12 3 ‐2 ‐1 0 1 4 ‐3 ‐2 ‐1 0 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 2) Suppose you have two regular six sided dice and you roll them together. If the value of each die is even, you add their values. If the value of each die is odd, you subtract the smaller value from the larger value, and if one die is odd and the other even, you multiply their values. a)
b)
c)
d)
e)
f)
Event B : result is greater than or equal to 12 Event C: result has 5 as a factor (divisible by 5) Event D: result is 8 Write out the sample space of this event. What is the P(B)? 9/36 = 1/4 What is the P(C)? 8/36 = 2/9 What is the P(D)? 3/36 = 1/12 What is P(B or C)? P(B) + P(C) – P(B∩C) = ¼ + 2/9 – 1/9 = 13/36 What is P(C ∩ D)? Ø 1
2
3
4
5
6
1 2 3 4 5 6 0 2 2 4 4 6 2 4 6 6 10 8 2 6 0 12 2 18 4 6 12 8 20 10 4 10 2 20 0 30 6 8 18 10 30 12 1
g) What is P(C | B)? P( B ∩ C ) = 9 = 4 P( B)
1
4
9
h) Are B and C independent? Why? No, The P(C) = 2/9 is not equal to P(C|B)= 4/9. So, C is not independent of B 3) Joyce has a 0.5 chance of getting a federal grant for college, and a 1/3 chance of getting a state grant. If the federal and state grants have no effect on one another. What is the probability that Joyce will get either the federal or the state grant? P(Federal) + P(State) – P(Federal∩State). Since Federal and State have no effect on each other, they are independent. Therefore, P(Federal∩State) = P(Federal)P(State)= ½*1/3= 1/6. Now, substituting into P(Federal) + P(State) – P(Federal∩State). ½+1/3‐1/6= 4/6 = 2/3 4) The following were the make and color of cars in a user automobile lot not far from school. Honda Toyota Mazda Black 4 3 13 20 Silver 16 10 14 40 White 8 12 5 25 Red 10 2 3 15 38 27 35 100 a) Fill in the marginal distributions b) What is the probability that the next car sold is red? 15/100 = 15% What is the probability that the next car sold is HondaC ? 1 – 38/100 = 62% What is the probability that the next car sold is a Mazda? 35/100 = 35% What is the probability that the next car sold would be a silver Toyota? 10/100 = 10% What’s the probability that the next car sold is neither black nor a Toyota? (16+8+10+14+5+3)/100 = 56/100 =56% g) If the next car sold is white, what’s the probability that it is a Honda? c)
d)
e)
f)
8
P( white ∩ Honda) 100 8
P( Honda | white) =
=
=
25 25
P( white)
100
h) What is the probability that the next car sold is a Madza, given that it’s black? 13
P(black ∩ Mazda) 100 13
P( Mazda | black ) =
=
=
20 20
P(black )
100
i)
What is the probability that the next car sold is a Toyota, given that it’s not black? 10 + 12 + 2
C
(
∩
)
24
P
black
Toyota
100
=
=
P(Toyota | black C ) =
C
20
80
P(black )
1−
100
5) There is a 32% chance Michael has bad breath and a 28% that he has BO (Body Odor). There is a 16% chance that he has both bad breath and body odor. a) Draw a Venn Diagram of the above scenario b) What is the probability that Michael has bad breath but not BO? P(Bad breath ∩ BOC) = 32%‐16% ‐ 16% c) What is the probability that Michael has either bad breath or body odor? P(Bad breath) + P(BO) – P(Bad breath ∩ BOC)= .32+.28 ‐ .16 = .44 d) What is the probability that Michael has BO given that he has bad breath? P ( BO | bad breath) =
P ( BO ∩ bad breath) .16 1
=
= P (bad breath)
.32 2
e) Are the events, bad breath and body odor mutually exclusive? Explain. No they share outcomes in common (there is overlap) f) Are the events, bad breath and body odor independent? Explain. The probability that Michael has BO is 28%. If BO and Bad Breath were independent, then Michael would have to have BO 28% of the time given he has bad breath, but the P(BO|Bad Breath) = 50%. Since these are not equal, these events are dependent (not independent). 6) Tiffany has a few choices after school. She can hang with her friends, do homework or go shopping. If she hangs with her friends she has a 70% chance of being happy. If she does homework, she only has a 40% chance of being happy. Finally, if she goes shopping she has an 85% chance of being happy. Typically, Tiffany does homework 55% of the time, hangs with her friends 35% of the time and shops 10% of the time. a) Draw a tree diagram that describes this problem Homework 55%
Tiffany
Friends 35%
Shopping 10%
Happy 40%
P(Homework∩Happy) = (.55)(.40) = .22 = 22% Not happy 60%
P(Homework∩HappyC) = (.55)(.60) = .22 = 33% Happy 70%
P(Friends∩Happy) = (.35)(.70) = .245 = 24.5% Not Happy 30%
P(Friends∩HappyC) = (.35)(.30) = .105 = 10.5% Happy 85%
P(Shopping ∩Happy) = (.10)(.85) = .085 = 8.5% P(Shopping∩HappyC) = (.10)(.15) = .015 = 1.5% b) What is the probability that Tiffany is happy? The probability that Tiffany will be happy is the sum of the probabilities that include “Happy”. P(Happy) = .22 + .245 + .085 = .55 = 55% c) If we have a happy Tiffany, what is the probability that she did homework? Not Happy 15%
P (homework | happy ) =
P (happy ∩ homework) .22 2
=
= = 40% .55 5
P (happy )
7) In a recent survey of Galileo Teachers college experiences, the following data was recorded (fictitious of course) Drank beer Did not drink regularly beer regularly Smoked 13 20 33 77 Did not 45 32 smoke (Mr. Page was in this category) 58 52 110 a) Write in the marginal distributions b) What’s the probability of a Galileo Teacher never smoking in college? 77/110 = 70% c) What’s the probability that a Galileo teacher was a partier (smoked and drank)? 13/110 = 11% d) What’s the probability that a Galileo Teacher was like Mr. Page (did not smoke or drink)? 32/110 = 29% e) If one of Galileo’s Teachers was a drinker, what’s the probability that they smoked too? 13
P (smoker ∩ drinker) 110 13
P (smoker | drinker) =
=
=
= 22.4% 58 58
P (drinker)
110
f)
Given that a Galileo Teacher was a smoker, what’s the probability that they were a drinker? 13
P (smoker ∩ drinker) 110 13
P (drinker | smoker) =
=
=
= 39.4% 33 33
P (smoker)
110
g) Are smoking and drinking independent? Explain your answer. P(drinking) = 58/110 = 52.7%. P(drinking|smoking) = 39.4%. Since these probability are not the same, smoking and drinking are NOT independent. 8) Elmer (AKA Lover‐Boy) takes his girlfriend to Great America. They walk by the games and she begs him to win her a huge stuffed animal. Elmer guesses that his probability to win one of the stuffed animals is 3% on any one try of the games. If Elmer’s guess is correct and he plays the game 20 times, what the probability that he wins her at least one stuff animal? P(at least one) = 1 – P(none) = 1 ‐ .9720 = 1 ‐ .5438 = 45.62% 9) Mr. Page is always running late but makes up time by driving fast. He needs to get to a special event on time, but of course he is leaving his house late. There are three ways to get to his destination, but he needs to go over the speed limit to get there on time. If he takes Route A he has a 25% chance of getting there on time. If he takes Route B, he has a 35% chance of getting there on time, and if he takes Route C, he has a 40% chance of getting there on time. Taking Route A, he has a 10% chance of getting a ticket. Taking Route B would increase his chance of getting a ticket to 15% and Route C chance of getting a ticket would be 25%. Ticket A
10%
25%
TicketC
90%
Ticket
Route
B
15%
35%
TicketC
85%
Ticket
C
25%
40%
TicketC
P(A∩Ticket) = (.25)(.10) = .025 = 2.5% P(A∩TicketC) = (.25)(.90) = .225 = 22.5% P(B∩Ticket) = (.35)(.15) = .0525 = 5.25% P(B∩TicketC) = (.35)(.85) = .2975 = 29.75% P(C∩Ticket) = (.40)(.25) = .10 = 10%
P(C∩TicketC) = (.40)(.75) = .30 = 30%
a) Draw a tree diagram of this problem b) What’s the probability that Mr. Page gets there on time without a ticket? This is all of the probabilities that include TicketC. P(TicketC) = .225 + .2975 + .30 = .8225 = 82.25% c) If Mr. Page didn’t get a ticket, what’s the probability he took the safest route (Route A)? 75%
P ( A | ticket C ) =
P (A ∩ ticket C ) .225
=
= .2735 = 27.35% P (ticket C )
.8225
d) If Mr. Page did get a ticket, what’s the likelihood he took the riskiest route (Route C)? P (C | ticket) =
.10
P (C ∩ ticket)
=
= .5634 = 56.34% 1 − .8225
P (ticket)