A 6 sided die is weighted so that it is 3 times more likely to land on the number 1 than 2. All other numbers, 2,3,4,5,6 have an equal chance of being rolled. If the die is thrown twice, what is the probability that the sum of the numbers will be 2?
A 6 sided die is weighted so that it is 3 times more likely to land on the number 1 than 2. All other numbers, 2,3,4,5,6 have an equal chance of being rolled. If the die is thrown twice, what is the probability that the sum of the numbers will be 2?
Hello, katinthehat108!
A 6-sided die is weighted so that it is 3 times more likely to land on the number 1 than 2.
All other numbers, 2,3,4,5,6 have an equal chance of being rolled.
If the die is thrown twice, what is the probability that the sum of the numbers will be 2?
The ratio of the probabilities of the numbers look like this:
. . $\displaystyle \begin{array}{|c||c|c|c|c|c|c|} \hline
\text{Number} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\text{Prob.} & 3 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
Hence, the probabilities are: .$\displaystyle P(1) = \frac{3}{8},\quad P(2) \,=\, P(3) \,=\, P(4) \,=\, P(5) \,=\, P(6) \,=\,\frac{1}{8}$
Therefore: .$\displaystyle P(\text{sum is 2}) \:=\:P(\text{1 and 1}) \:=\:\frac{3}{8}\cdot\frac{3}{8} \:=\:\frac{9}{64}$