2
$\begingroup$

Are the following probabilities correct? I'm not very confident with probabilities and would just like these double checked please. Thank you.

Experiment: Roll three 6-sided dice.

a) Find the probability that none of the dice rolled are a 2.

$\left(\frac{5}{6}\right)^3$

b) Find the probability that the sum of the dice is 8.

$\frac{21}{216}$

c) Find the probability that the sum of the dice is 8 and none of the dice rolled are 2’s.

$\frac{9}{216}$

d) Find the probability that at least one of the dice rolled are a 2.

$1- \left(\frac{5}{6}\right)^3$

e) Find the probability that the three values are sequential (for example: 4, 3, 5.)

$1 \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{4}{6}\right)$

f) Find the probability that none of the dice rolled are a 2 given that the sum is 8.

$\frac{9}{21}$

g) Find the probability that the sum is 8 given that none of the dice rolled are a 2.

$\frac{9}{125}$

h) Are the events “none of the dice rolled are 2” and “the sum is 8” independent? Explain.

Yes, because they don't depend on each other. For example, you can roll a sum of 8 while still rolling a 2 or without a 2.

i) Assume that the dice are colored: one is red and two are blue. Find the probability that the sum of the blue dice is equal to the value on the red die.

$\frac{15}{216}$

j) Assume that the dice are colored: one is red and two are blue. Find the probability that the sum of the blue dice is equal to the value on the red die given that the red value is 6.

$\frac{5}{36}$

k) Assume that the dice are colored: one is red and two are blue. Find the probability that the value of the red die is 6 given that the sum of the blue dice is equal to the value on the red die.

$\frac{5}{15}$

$\endgroup$
2
  • $\begingroup$ If no one else does comment me and I'll help you, but the first one is right,, showing your workings would help. Looks like you get the logic! $\endgroup$
    – Alec Teal
    May 1 '15 at 13:12
  • $\begingroup$ In order to establish independence, you must show that $P(A \cap B) = P(A)P(B)$. $\endgroup$ May 1 '15 at 16:08
2
$\begingroup$

Most of your work is correct.

There are two exceptions.

If the three dice are sequential, then they must assume one of the following sets of values $\{1, 2, 3\}$, $\{2, 3, 4\}$, $\{3, 4, 5\}$, or $\{4, 5, 6\}$. For each of the four sets, there are $3! = 6$ permutations which result in the same set of values. Thus, the probability that the dice are sequential is $$\frac{4 \cdot 3!}{6^3} = \frac{24}{216} = \frac{1}{9}$$

If events $A$ and $B$ are independent, then $P(A \cap B) = P(A)P(B)$. If we let $A$ be the event that the sum of the dice is $8$ and $B$ be the event that none of the dice rolled show a 2, then \begin{align*} P(A) & = \frac{21}{216} = \frac{7}{72}\\ P(B) & = \frac{125}{216}\\ P(A \cap B) & = \frac{9}{216} = \frac{3}{72} \end{align*} As you can check, $P(A \cap B) \neq P(A)P(B)$. The events are dependent because if none of the dice rolled show a 2, you are less likely to get a sum of $8$ since it becomes more likely that the sum of the numbers shown on the dice will be too large.

$\endgroup$
1
$\begingroup$

Your answers all look correct except for (e) and (h). On (e), since you know that there are $6^3 = 216$ different rolls of the three dice, simply count up the rolls in which the three numbers are consecutive. There are four such selections: $\{1, 2, 3\}, \{2, 3, 4\}, \{3, 4, 5\}, \{4, 5, 6\}$. Each selection can be ordered in $3! = 6$ different ways. There are thus $4 \cdot 6 = 24$ different rolls that satisfy the condition, and the probability is therefore $24/216 = 1/9$.

As N. F. Taussig points out, you are applying too weak a condition for independence on part (h). It is not sufficient merely that event $A$ can happen regardless of event $B$, and vice versa; each one's probability of occurrence must be unaffected by the other's occurrence (or non-occurrence). The most straightforward way to determine that is to verify that $P(A \cap B) = P(A)P(B)$. You have already determined the relevant quantities in parts (a), (b), and (c); is the product rule satisfied?

$\endgroup$

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.