203,390 views
41 votes
41 votes
A fair Octahedral (eight-sided) die has the

following numbers on te's aces 2,3,1,2,13,11.
each time the die is thrown the score is the
humber on the top face if the die is thrown.
a Once, what is the probability that the scorce is
not 3
b twice, find the probability of scoring a sum
twice, find the probability of Scoring a sum
of 5​

User Sajith Herath
by
3.0k points

2 Answers

24 votes
24 votes

Final answer:

The probability that the score is not 3 upon rolling the eight-sided die once is 7/8. To find the probability of scoring a sum of 5 when rolling the die twice, count the combinations that sum up to 5 and divide by the total number of outcomes, which is 64.

Step-by-step explanation:

The question involves calculating the probability of various outcomes when rolling an eight-sided die with the faces 2, 3, 1, 2, 13, 11. The first part of the question asks for the probability that the score is not 3. Since there are eight sides and only one face has the number 3, there are seven outcomes where the score is not 3. Therefore, the probability is 7/8.

The second part of the question asks for the probability of scoring a sum of 5 when rolling the die twice. The sample space for two throws has 64 outcomes (8x8). There are a few combinations of throws that result in a sum of 5, such as (2, 3) and (1, 4) if a '1' and '4' are available on the die, but since the faces are not specified for all numbers, we cannot give all combinations. Assuming only the mentioned numbers are on the die faces, we must look for combinations that add up to 5, count them, and then divide by 64 to find the probability.

User Uan
by
3.2k points
18 votes
18 votes

Answer:

30%

Step-by-step explanation:

Your chance of scoring a 5 is 30%

User Chansik Im
by
2.5k points
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