A six-sided number cube is weighted so that the probabilities of throwing 2, 3, 4, 5, or 6 are equal, and the probability of throwing a 1 is twice the probability of throwing a 2. if the number cube is thrown twice, what is the probability that the sum of the numbers thrown will be 4?

Question

Bridger Mclean

Mathematics

26 December, 06:22

A six-sided number cube is weighted so that the probabilities of throwing 2, 3, 4, 5, or 6 are equal, and the probability of throwing a 1 is twice the probability of throwing a 2. if the number cube is thrown twice, what is the probability that the sum of the numbers thrown will be 4?

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Dashawn Lee · Answer 1

Answer: The probability of rolling 4 in this scenario would be 5/49.

First, lets identify the probabilities of each event. Since, rolling a 1 is twice the probability as the other events, rolling a 1 would be 2/7 and the rest would be 1/7. Adding up these probabilities would give a total of 1 whole.

Second, let find all the possibilities of rolling 4 and add up their probabilities.

P (2, 2) = 1/7 x 1/7 = 1/49

P (1, 3) = 2/7 x 1/7 = 2/49

P (3, 1) = 1/7 x 2/7 = 2/49

Add up those fractions and you will have 5/49.