Probability answers

1. I have two 10-sided dice, numbered 1-10. If I toss both of them, the sums are shown on this chart:

One place to start is to make a grid: (grey is numbers on the dice)

(yes, I know this is a pretty big grid. I put it on the practice problems so that I could save the smaller grid problems for the quiz)

And mark evens, numbers over 10, equal to 10 and doubles (see graphics). Then count to find the probabilities

• a sum that is an even number? 50/100=1/2
• a sum that is over 10? 55/100=11/20
• a sum that is equal to 10? 9/100
• doubles? 10/100=1/10

2. I have two funny spotted 6-sided dice. Die A has the numbers: 1, 1, 3, 3, 5, 6 and die B has the numbers 2, 2, 2, 4, 4, 4

We can make a grid here too; in this grid I am marking which die would be larger:

• which die (if any) will be larger most often? Neither, they are each largest half of the time.
• What is the probability that the first die will be bigger than the second? 1/2

3. If I have the pair of spinners below, what is the probability, when I spin both, that the sum will be 6?

And yet another grid:

P(3) = 3/12=1/4 (the probability of getting a sum of 3: see yellows)
P(4) = 4/12=1/3 (the probability of getting a sum of 4: see blues)

4. If I have a bag with 5 red and 3 blue marbles, and I pick two out (one and then the other,without replacing the first)

These are best done by multiplication:

• What is the probability I will get two red marbles? (5/8)(4/7)=20/56=5/14

  Note that 5/8 is the probability of getting a red the first time, after which there are 4 red out of 7 total marbles left, for a probability of 4/7

• What is the probability I will get two blue marbles? (3/8)(2/7)=6/56=3/28
• What is the probability that I will get one red and one blue?

  You can do this two ways: add together red then blue and blue then red: (5/8)(3/7)+(3/8)(5/7)=(15/56)+(15/56)=30/56=15/28
  OR you can subtract the probabilities of RR and BB from 1 (since one of each is the only other possible outcome):
  1-(5/14)-(15/28)=(56/56)-(20/56)-(6/56)=30/56=15/28

5. Each of the spinners below costs $4.00 to play. If your friend is planning to play 100 times, which spinner would you reccommend that he use?

If he used spinner a he would get $5 about 25 times, $4 about 50 times, and $3 about 25 times; so he would get about 5*25+4*50+3*25=$400, and he would pay $400, so he would probably break even.

If he used spinner b, he would get $0 about 20 times, $5 about 40 times, $2 about 20 times and $6 about 20 times, so he would get about: 0*20+5*40+2*20+6*20=$360, so after paying $400, he would probably lose about $40.

If he used spinner c, he would get $6 about 50 times, $2 about 25 times and $3 about 25 times, so he would get about 6*50+2*25+3*25=$425, so after paying $400, he would probably win about $25.

If your friend were playing 100 times, it would be smartest to play game c.

6. Mystico the astrologer for a well read newspaper predicted that if you are an Aquarius, you will soon hear from a friend you have not spoken to in a very long time. For the purposes of this assignment, assume that "a very long time" means more than 10 years, and that "soon" means this week.
A. Is it likely or unlikely that this will happen to every Aquarius who is over age 15? (Explain)

Not very likely--it certainly doesn't happen to most people every week

B. Is it likely or unlikely that this will happen to at least one Aquarius who reads Mystico’s column? (Explain)

Yes, if lots of people read the column (say a million or two), it's almost certain that at least one of them will hear from an old friend during the week.

7. If there is a 40% chance that it will rain, there is a 60% chance that it will not rain (100%-40%=60%)