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

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- Sep 5th 2007, 01:37 AMprocrastinatorProbability 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 - Sep 5th 2007, 04:01 AMSoroban
Hello, procrastinator!

With "dice" problems, there are no neat formulas.

Usually, we must list the outcomes (or visualize them).

Quote:

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:

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