SOLUTION: a six sided die is weighed so that the probability of "4" is 4 times as likely as getting any other number. what is the probability of getting a "1" or a "3"?

Question 238130: a six sided die is weighed so that the probability of "4" is 4 times as likely as getting any other number. what is the probability of getting a "1" or a "3"?
Found 2 solutions by RAY100, solver91311:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
with a "normal" die,,,each side has equal chance, with 6 total possibilities.
With the weighted die,,#4,,, has 4 chances,,making the total 4+5=9
.
All of the unweighted sides have a probability of 1/9 .
.
The weighted side has a probability of 4/9.
.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

A fair 6-sided die would have a 1 in 6 probability of a 4. But this die has a 4 of something probability when each of the other possibilities has a 1 in something probability. Since the numerator of the "get a 4" probability is 4 and the other 5 are 1, we have a denominator of 9: 4 plus 5 times 1 = 9.

The probability of getting a 1 or a 3 is the sum of the probability of getting a 1 plus the probability of getting a 3. 1/9 + 1/9 = 2/9

The only thing that bothers me about this problem is how someone would weight a die so that the 4 was equally more likely than any of the other numbers. To make the 4 come up more often, you would have to weight the side with the 3, certainly making the 4 more likely, but also making the 3 somewhat less likely than the 1, 2, 5, or 6. I guess this is the same as "neglecting air resistance" when you are asked to calculate the height of a thrown rock.

John

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