Problems about experimental probabilities:

1. I dropped a lot of cans on the floor. Some of them landed on their side, and some of them landed on their end. Use this table to find the experimental probability of a can landing on its side and the experimental probability of a can landing on its end:

Landed on side: 26
Landed on end: 12

(give your answer as a decimal with 2 decimal places)

Total number of cans dropped = 26+12=38

P(land on side)= 26/38 = .68

P(land on end) = 12/38 = .32

Problems about complements:

2. a. If I spin 25 pennies, and 14 land heads up, what percentage of heads did I get? 56% (14 divided by 25 is .56)

b. What percentage tails did I get? (1-.56=.44) 44%

3. a. If I roll a normal die 20 times, and I get "6" 5 times, what percentage of the rolls were 6? (5/20=.25) 25%

b. What percentage of the rolls were less than 6? (100%-25%=75%) 75%

4. I have some red, blue and green beads in a bag. I did the following experiment 20 times: I took out three beads, and recorded if at least two were the same color. 70% of the time at least two were the same color. What happened the other 30% of the time? The other 30% of the time, they were all 3 different colors

5. I rolled two dice, and recorded whether I got the same number on both dice or not. 15% of the time, I got the same number on both dice. What happened the other 85% of the time? The other 85% of the time they were different numbers.

Problems about rolling multiple dice:

10. If I roll two 8-sided dice , what is the probability that I get two different numbers?

For the first die, anything I roll can give me different numbers, so I have 8/8=1 for the first die
For the second die, I have to roll something different from the first die. There are 7 numbers that are different from what was rolled by the first die, so the probability for the second die is 7/8.
Multiply these together P(different numbers) = 1*7/8 = 7/8

11. If I roll two 8-sided dice, what is the probability that I roll doubles?

I have two ways to do this:

First way:

For the first die, anything I roll can give me doubles, so I have 8/8=1 for the first die.
For the second die, I have to get the same number as on the first die. The second die only has 1 number that it can roll to make it doubles, so the probability for the second die is 1/8
Multiply these together: P(doubles) = 1*1/8 = 1/8

Second way:

If I don't roll different numbers that means that both numbers are the same, so
P(doubles) = 1-P(different) = 1-7/8 = 1/8

12. If I roll three 8 sided dice, what is the probability that I get triples (all 3 the same)?

For the first die, anything I roll can give me triples, so I have 8/8=1 for the first die.
For the second die, in order to get triples, I have to roll the same number that I rolled on the first die. There's only one number that will do that, so the probability for the second die is 1/8.
For the fhird die, in order to get triples, I have to roll the same number that I rolled on the first and the second roll. There's only one number that will do that, so the probability for the second die is also 1/8.
Multiply those all together to get the probability of triples: 1*1/8*1/8=1/64

13. If I roll three 8-sided dice, what is the probability that I get 3 different numbers?

For the first die, anything I roll can give me different numbers, so I have 8/8=1 for the first die
For the second die, I have to roll something different from the first die. There are 7 numbers that are different from what was rolled by the first die, so the probability for the second die is 7/8.
For the third die, I have to roll something different from the numbers on the first two dice. The first two dice took 2 numbers, so there are 6 numbers left that are different, so the probability for the third die is 6/8.
Multiply these together P(different numbers) = 1*7/8*6/8 = 42/64 = 21/32

14. If I roll three 8-sided dice, what is the probability that I get doubles (2 the same, and the third different)?

There are a lot of ways I could get doubles: the first two could be the same, and the third die could be different, or the first and third could be the same, or the second and third could be the same. That would be pretty confusing. So, we'll do it using complements:

If you roll 3 dice, and they aren't all the same and they aren't all different, then two have to be the same and one different. So, the probability of getting doubles is 1 (100%) - the probability of getting all the same - the probability of getting all different:

P(doubles) = 1 - P(triples) - P(different) = 1 - 1/64 - 42/64 = 64/64 - 43/64 = 21/64