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Chapter 2: Problem 10

Suppose three fair dice are rolled. What is the probability at most one six appears?

Short Answer

Expert verified
The probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).

Step by step solution

01

Define events and complementary events

Let A be the event "at most one six appears." We are asked to find P(A). Let A' be the event "at least two sixes appear" (the complementary event). Then, we can express A as the complement of A': P(A) = 1 - P(A').
02

Calculate the probability of each outcome for a single di茅

Since each die is fair and has six sides, the probability of rolling a six is 1/6 and the probability of not rolling a six is 5/6.
03

Compute the probability of A' using combinations

A' can occur when either two or three sixes appear. We can calculate the probability of these situations and sum them: a) Two sixes and one non-six: There are \(\binom{3}{2}\) = 3 ways to choose which two dice will have the sixes. The probability of this configuration is \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\). b) All three sixes: There is only one way that all three dice can have sixes, which has probability \(\left(\frac{1}{6}\right)^3\).
04

Compute P(A')

Now let's sum the probabilities from Step 3 to find P(A'): P(A') = 3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\).
05

Compute P(A) using the complement

Now that we have P(A'), we can compute P(A) using the complement relationship: P(A) = 1 - P(A') = 1 - [3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\)].
06

Evaluate and simplify

Evaluate and simplify the expression from Step 5: P(A) = 1 - [3 * \(\left(\frac{1}{36}\right)\left(\frac{5}{6}\right)\) + \(\frac{1}{216}\)] = 1 - [\(\frac{5}{72}\) + \(\frac{1}{216}\)]. First, convert the fractions to a common denominator (the least common multiple of 72 and 216 is 216): P(A) = 1 - [\(\frac{15}{216}\) + \(\frac{1}{216}\)]. Now, we can simplify the expression by combining the fractions: P(A) = 1 - \(\frac{16}{216}\). Simplify further by dividing by the greatest common divisor (8): P(A) = 1 - \(\frac{2}{27}\). Finally, convert the mixed number into an improper fraction: P(A) = \(\frac{27 - 2}{27}\) = \(\frac{25}{27}\).
07

State the final answer

So, the probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).

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Most popular questions from this chapter

Suppose that an experiment can result in one of \(r\) possible outcomes, the \(i\) th outcome having probability \(p_{i}, i=1, \ldots, r, \sum_{i=1}^{r} p_{i}=1 .\) If \(n\) of these experiments are performed, and if the outcome of any one of the \(n\) does not affect the outcome of the other \(n-1\) experiments, then show that the probability that the first outcome appears \(x_{1}\) times, the second \(x_{2}\) times, and the \(r\) th \(x_{r}\) times is $$ \frac{n !}{x_{1} ! x_{2} ! \ldots x_{r} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{r}^{x_{r}} \quad \text { when } x_{1}+x_{2}+\cdots+x_{r}=n $$ This is known as the multinomial distribution.

Let \(X\) be a random variable with probability density $$ f(x)=\left\\{\begin{array}{ll} c\left(1-x^{2}\right), & -1

Suppose that \(X\) takes on each of the values \(1,2,3\) with probability \(\frac{1}{3} .\) What is the moment generating function? Derive \(E[X], E\left[X^{2}\right]\), and \(E\left[X^{3}\right]\) by differentiating the moment generating function and then compare the obtained result with a direct derivation of these moments.

Let \(X\) be binomially distributed with parameters \(n\) and \(p\). Show that as \(k\) goes from 0 to \(n, P(X=k)\) increases monotonically, then decreases monotonically reaching its largest value (a) in the case that \((n+1) p\) is an integer, when \(k\) equals either \((n+1) p-1\) or \((n+1) p\), (b) in the case that \((n+1) p\) is not an integer, when \(k\) satisfies \((n+1) p-1<\) \(k<(n+1) p\) Hint: Consider \(P\\{X=k\\} / P\\{X=k-1\\}\) and see for what values of \(k\) it is greater or less than 1 .

Suppose \(X\) has a binomial distribution with parameters 6 and \(\frac{1}{2}\). Show that \(X=3\) is the most likely outcome.
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