Hi >.<
I got another Q:
The number showing on the upper faces when 2 fair six-sided dice are tossed, is observed. Consider events A and B
A- the total score is at most 5
B- at least one die achieves a score of 3 or 4
thanks
Hello, procrastinator!
With "dice" problems, there are no neat formulas.
Usually, we must list the outcomes (or visualize them).
I assume we are to find $\displaystyle P(A)$ and $\displaystyle P(B)$.The number showing on the upper faces when two fair six-sided dice are tossed, is observed.
Consider events A and B
A: the total score is at most 5
B: at least one die achieves a score of 3 or 4
(1) There are: .$\displaystyle 6 \times 6 \:=\:36$ possible outcomes.
There are 10 outcomes with a sum of 5 or less:
. . $\displaystyle (1,1),\,(1,2),\,(1,3),\,(1,4),\,(2,1),\,(2,2),\,(2 ,3),\,(3,1),\,(3,2),\,(4,1)$
Therefore: .$\displaystyle P(A) \;=\;\frac{10}{36} \;=\;\frac{5}{18}$
(2) I have a "back door" approach to the second part.
The probability that a die does not show a 3 or 4 is: .$\displaystyle \frac{4}{6} \,=\,\frac{2}{3}$
The probability that neither die shows a 3 or 4 is: .$\displaystyle \frac{2}{3}\!\cdot\!\frac{2}{3}\:=\:\frac{4}{9}$
Therefore, the probability that at least one die shows a 3 or 4 is:
. . $\displaystyle P(B) \;=\;1 - \frac{4}{9}\;=\;\frac{5}{9}$