# Thread: Probability Q Again..

1. ## Probability Q Again..

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

2. Hello, procrastinator!

With "dice" problems, there are no neat formulas.
Usually, we must list the outcomes (or visualize them).

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
I assume we are to find $\displaystyle P(A)$ and $\displaystyle P(B)$.

(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:

. . $\displaystyle P(B) \;=\;1 - \frac{4}{9}\;=\;\frac{5}{9}$