0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How do you find the P(not a five) for any of the dice $\endgroup$ – Satish Ramanathan Oct 31 '13 at 18:15
  • $\begingroup$ For any event $E$, $P(E^c) = 1-P(E)$. So, $P(\text{not a five}) = 1 - P(\text{is a five})$. $\endgroup$ – Tom Oct 31 '13 at 18:16
0
$\begingroup$

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:

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

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

$$= 0.40136.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.