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An experiment consists of $3$ fair, different coloured dice being rolled. The dice are $6$-sided and the sides show numbers $1,\dots,6$. Let $A$ be the event that none of the dice shows numbers $1$ and $2$, and let $B$ be the event that all dice show an odd number.

A) What is the probability of $A$?

B) What is the probability of $B$?

C) What is the probability of $A$ intersecting $B$?

I've solved this question by finding the total number of possible outcomes: $|S| = 6^3 = 216$

The results were way too long for the marks given which makes me question the method I used for these solutions. I ended up with:

A) $P(A) = \frac{64}{216} = \frac{8}{27} $

B) $P(B) = \frac{26}{216} = \frac{13}{108}$

C) $P(A \cap B) = \frac{8}{216} = \frac{1}{27}$

T. Mike
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3 Answers3

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$P(A)=\frac{2}{3}^3=\frac{8}{27}$ You are correct.

$P(B)=\frac{1}{2}^3=\frac{1}{8}$ You are incorrect.

$P(C)=\frac{1}{3}^3=\frac{1}{27}$ ou are correct.

herb steinberg
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  • Thank you! But can you please explain how you ended with your results because I'm looking for a simpler way of finding the solutions – T. Mike May 11 '19 at 20:18
  • Since the dice are independent the cube for all answers results from getting the answer for one die and using the basic formula that the probability of several independent events is the product of the probabilities of the individual events. For (A) there are 4 possibilities for the event (3,4,5,6) out of 6 total For (B) there are 3 possibilities (1,3,5) out of 6 total. For (C) there are 2 possibilities (3,5) out of 6 total. – herb steinberg May 11 '19 at 21:32
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So to think about $A$ for the first role we have $\frac{4}{6}$ possibility as we only want $3,4,5,6$ and this is the same for the other $3$ dices, so $(\frac{4}{6})^3= \frac{8}{27}$.

Next for $B$ we have have the odd numbers to be $1,3,5$ so the possibilities are $\frac{3}{6}$ for each dice, so overall again we cube for three dices so $(\frac{3}{6})^3= \frac{1}{8}$.

Finally we have odd and not one or two, so the number on one dice can only be $3,5$ so with 3 dices: $(\frac{2}{6})^3= \frac{1}{27}$.

But your answers are correct hope this helps :)

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A) The number of ways $3$ dice can not show a $1$ or $2$ is $4\cdot4\cdot4 = 64$. The total number of outcomes is $6\cdot6\cdot6 = 216$

The probability is therefore $\frac{64}{216} = \frac{8}{27}$.

B) The number of ways $3$ dice can show an odd number is $3\cdot3\cdot3 = 27$. Like before, the total number of outcomes is $216$.

The probability is therefore $\frac{27}{216} = \frac{1}{8}$

C) The number of ways $3$ dice cannot be a 1 or 2 or an even number is $2\cdot2\cdot2 = 8$.

The probability is therefore $\frac{8}{216} = \frac{1}{27}$.

Phil H
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