# What is the probability that the number $5$ comes up on exactly two of three loaded dice?

I roll three different loaded dice. For the first die, the probability of getting a $$5$$ is $$0.7$$, for the second die the probability of getting a $$5$$ is $$0.48$$, and for the third die the probability of getting a $$5$$ is $$0.38$$.

What is the probability that the number $$5$$ comes up on exactly two of the three dice?

If $E_i$ is the event that dice $i$ ($i=1,2,3$) comes up a $5$, then the event of getting exactly two 5s is $$\big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big)$$ Each of the pieces in parentheses is disjoint from the others (why!?) so using the properties of the probability $P$, $$P(\text{exactly two } 5s) = P\big\{\big(E_1 \cap E_2 \cap E_3^c\big)\cup\big(E_1 \cap E_2^c \cap E_3\big)\cup(E_1^c \cap E_2 \cap E_3\big)\big\} = P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big).$$ Now, assuming that the $E_i$ are independent, $$P\big(E_1 \cap E_2 \cap E_3^c\big) + P\big(E_1 \cap E_2^c \cap E_3\big)+P\big(E_1^c \cap E_2 \cap E_3\big) = P(E_1)P(E_2)P(E_3^c) + P(E_1)P(E_2^c)P(E_3) +P(E_1^c)P(E_2)P(E_3)$$ and now you can figure out the right side of the last equality with the information you're given.

• How do you find the P(not a five) for any of the dice – Satish Ramanathan Oct 31 '13 at 18:15
• For any event $E$, $P(E^c) = 1-P(E)$. So, $P(\text{not a five}) = 1 - P(\text{is a five})$. – Tom Oct 31 '13 at 18:16

Because the probability of each die coming up $$5$$ is different, we can't short-cut this by using combinatorics. We have to consider all the possibilities.

Let's call the probability that a die comes up $$5$$ "$$F$$", and not-$$5$$ "$$G$$". Then, let's label each of the dice with a suffix of $$1$$, $$2$$, or $$3$$.

Of course, the $$G$$ value for each die is $$1$$ minus the $$F$$ value.

As you said "exactly $$2$$", then we must consider all the combinations with exactly $$2$$ $$F$$'s and $$1$$ $$G$$. Namely:

$$P = F1 \times F2 \times G3 + F1 \times G2 \times F3 + G1 \times F2 \times F3$$

$$= 0.7 \times 0.48 \times (1-0.38) + 0.7 \times (1-0.48) \times .38 + (1-0.7) \times 0.48 \times 0.38$$

$$= 0.40136.$$