A four sided die and a six sided Die are thrown. The four sided die is weighted

Question

A four sided die and a six sided Die are thrown. The four sided die is weighted such that the probability that it shows 1 is 1/10, the probability that it shows 2 is 2/10, the probability that it shows 3 is 3/10 and the probability that it shows 4 is 4/10. The six sided die is a FAIR die.

What is the probability of getting a double?

What is the probability that the sum of the numbers shown on the dice is 5?

What is the probability that both dice show even numbers?

What is the probability that both dice show prime numbers? Note: 1 is not a prime number!

What is the probability that the sum of the numbers shown on the dice is at least 8?

Explanation / Answer

For the fair die, probability of showing any of the 6 numbers is 1/6

(a) Probability of getting a double = P(both 1's) + P(both 2's) + P(both 3's) + P(both 4's)

P(both 1's) = 1/10 * 1/6 = 1/60

P(both 2's) = 2/10 * 1/6 = 2/60

P(both 3's) = 3/10 * 1/6 = 3/60

P(both 4's) = 4/10 * 1/6 = 4/60

Therefore the required probability = (1/60) + (2/60) + (3/60) + (4/60) = 10/60 = 1/6

(b) P(Sum =5). The different possibilities are (1+4) (4 +1) (2 + 3) and (3 + 2) where the 1st number is on the 4 faced dice.

Probability of 1 and 4 = 1/10 * 1/6 = 1/60 and probability of 4 and 1 = 4/10 * 1/6 = 4/60

Probability of 2 and 3 = 2/10 * 1/6 = 2/60 and probability of 3 and 2 is = 3/10 * 1/6 = 3/60

Therefore the required probability = 1/60 + 4/60 + 2/60 + 3/60 = 10/60 = 1/6

(c) P(both dice show even numbers):The different possibilities are (first number is the four faced dice)

(2,2), (2,4), (2,6), (4,2), (4,4), (4,6)

For (2,2), (2,4), (2,6) each event has a probability = 2/10 * 1/6 = 2/60. Therefore for the 3 events = 6/60

For (4,2), (4,4), (4,6) each event has a probability = 4/10 * 1/6 = 4/60. Therefore for the 3 events = 12/60

The required probability = 6/60 + 12/60 = 18/60 = 3/10

(d) P(both dice show prime numbers):The different possibilities are (first number is the four faced dice)

(2,2) (2,3) (2,5), (3,2) (3,3) (3,5) (4,2) (4,3) (4,5)

For (2,2), (2,3), (2,5) each event has a probability = 2/10 * 1/6 = 2/60. Therefore for the 3 events = 6/60

For (3,2) (3,3) (3,5) each event has a probability = 3/10 * 1/6 = 3/60. Therefore for the 3 events = 9/60

For (4,2), (4,3), (4,5) each event has a probability = 4/10 * 1/6 = 4/60. Therefore for the 3 events = 12/60

The required probability = 6/60 + 9/60 + 12/60 = 27/60 = 9/20

(e) At least 8 means 8 or greater than 8: The different possibilities are (first number is the four faced dice)

(2,6), (3,5), (3,6), (4,4), (4,5) and (4,6)

For (2,6) the probability = 2/10 * 1/6 = 2/60

For (3,5), (3,6), each event has the probability = 3/10 * 1/6 = 3/60. Therefore for the 2 events = 6/60

For (4,4), (4,5), (4,6) each event has a probability = 4/10 * 1/6 = 4/60. Therefore for the 3 events = 12/60

The required probability = 2/60 + 6/60 + 12/60 = 20/60 = 1/3

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