Question

1. Suppose you are rolling a fair four-sided die and a fair six-sided die and you are counting the number of ones that come up.
   1. What is the probability that both die roll ones?
   2. What is the probability that exactly one die rolls a one?
   3. What is the probability that neither die rolls a one?
   4. What is the expected number of ones?
   5. If you did this 1000 times, approximately how many times would you expect that exactly one die would roll a one?
   6. What is the expected number of ones?

Answer 1

One die is of 4 sided and other is 6 sided.

a) P(Both 1s) = (1/4) * (1/6) = 1/24

b) P(Exactly one 1) = (1/4) * (5/6) + (3/4) * (1/6) = 8/24 = 1/3

c) P(None 1s) = (3/4) * (5/6) = 5/8

d) E(x) = 0*(5/8) + 1*(1/3) + 2*(1/24) = 5/12

e) 1000 times then: 1000*(1/3) = 1000/3 = 333.3

Hence, we expect 333 times exactly one 1s.

f) This is same as d)

E(x) = 0*(5/8) + 1*(1/3) + 2*(1/24) = 5/12

Answer 2

To find the probabilities, we need to consider the outcomes of rolling the four-sided die (D4) and the six-sided die (D6). Both dice are fair, meaning each outcome has an equal probability of 1/n, where n is the number of sides on the die.

Let's label the outcomes of rolling the D4 as 1, 2, 3, and 4, and the outcomes of rolling the D6 as 1, 2, 3, 4, 5, and 6.

(a) Probability that both dice roll ones: P(rolling a one on D4) = 1/4 P(rolling a one on D6) = 1/6

Since the dice are independent, the probability of both events occurring is the product of their individual probabilities: P(rolling ones on both dice) = P(rolling a one on D4) × P(rolling a one on D6) = (1/4) × (1/6) = 1/24

(b) Probability that exactly one die rolls a one: We have two scenarios for this: either D4 rolls a one and D6 doesn't, or D4 doesn't roll a one and D6 does.

P(D4 rolls a one and D6 doesn't) = P(rolling a one on D4) × P(not rolling a one on D6) = (1/4) × (5/6) = 5/24

P(D4 doesn't roll a one and D6 does) = P(not rolling a one on D4) × P(rolling a one on D6) = (3/4) × (1/6) = 3/24

Add the probabilities of the two scenarios: P(exactly one die rolls a one) = P(D4 rolls a one and D6 doesn't) + P(D4 doesn't roll a one and D6 does) = 5/24 + 3/24 = 8/24 = 1/3

(c) Probability that neither die rolls a one: P(not rolling a one on D4) = 3/4 P(not rolling a one on D6) = 5/6

The probability of both dice not rolling ones is the product of their individual probabilities: P(neither die rolls a one) = P(not rolling a one on D4) × P(not rolling a one on D6) = (3/4) × (5/6) = 15/24 = 5/8

(d) Expected number of ones: The expected number of ones is the sum of the probabilities of all the events where a one is rolled multiplied by the number of ones rolled in that event:

Expected number of ones = (1) × P(rolling ones on both dice) + (1) × P(exactly one die rolls a one) + (0) × P(neither die rolls a one) Expected number of ones = (1) × (1/24) + (1) × (1/3) + (0) × (5/8) = 1/24 + 1/3 + 0 = 1/24 + 8/24 = 9/24 = 3/8

(e) If you did this 1000 times, approximately how many times would you expect that exactly one die would roll a one? Since the probability of exactly one die rolling a one is 1/3, the expected number of times this event will occur in 1000 trials is: Expected number of times = (1/3) × 1000 = 333.33 (approximately 333 times).

Answer 3

Let's analyze the probabilities for each scenario:

1. Probability that both dice roll ones: The probability of rolling a one on a fair four-sided die is 1/4, and the probability of rolling a one on a fair six-sided die is 1/6. Since the dice rolls are independent events, the probability of both dice rolling ones is the product of their individual probabilities:

Probability = (1/4) * (1/6) = 1/24 ≈ 0.0417 or 4.17%

2. Probability that exactly one die rolls a one: To calculate this probability, we need to consider two cases: one die rolling a one while the other does not, or the other die rolling a one while the first one does not.

Probability = (1/4) * (5/6) + (3/4) * (1/6) = 5/24 ≈ 0.2083 or 20.83%

3. Probability that neither die rolls a one: This is the complement of the probability that at least one die rolls a one.

Probability = 1 - (5/24) = 19/24 ≈ 0.7917 or 79.17%

4. Expected number of ones in a single roll: To calculate the expected number of ones, we multiply each outcome by its probability and sum them up:

Expected number of ones = (1 * 1/24) + (1 * 5/24) + (0 * 19/24) = 6/24 = 1/4 ≈ 0.25

5. If you did this 1000 times, approximately how many times would you expect exactly one die to roll a one? To find the expected number of times exactly one die rolls a one in 1000 trials, we multiply the probability of exactly one die rolling a one by the number of trials:

Expected number of times = (5/24) * 1000 ≈ 208.33

Please note that the expected number of times is an average value, so in this case, you can expect approximately 208 times out of 1000 trials that exactly one die would roll a one.