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Chapter 2: Problem 18

Show that when \(r=2\) the multinomial reduces to the binomial.

Short Answer

Expert verified
When \(r=2\), the multinomial theorem becomes \(\begin{pmatrix}n\\ m_1, m_2 \end{pmatrix}=\frac{n!}{m_1!m_2!}\) with \(m_2=n-m_1\). Replacing \(m_2\) with \(n-m_1\) gives us \(\begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n-m_1)!}\), which is equivalent to the binomial theorem \(\binom{n}{m_1}=\frac{n!}{m_1!(n-m_1)!}\). Hence, when \(r=2\), the multinomial reduces to the binomial.

Step by step solution

01

Input \(r=2\) into the Multinomial Theorem

When we input \(r=2\) into the multinomial theorem, we have two types of events (\(m_1\) and \(m_2\)). The multinomial becomes: \[ \begin{pmatrix}n\\ m_1, m_2 \end{pmatrix}=\frac{n!}{m_1!m_2!}. \] Since we have only 2 types of events and the total number of events is \(n\), we can express the relationship between \(m_1\) and \(m_2\) as \[ m_2=n-m_1\]
02

Replace \(m_2\) with \(n - m_1\)

Now that we have expressed the relationship between \(m_1\) and \(m_2\), we will replace \(m_2\) with \(n - m_1\) in the multinomial: \[ \begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n-m_1)!}. \]
03

Show Equivalence to the Binomial Theorem

Now let's compare our reduced multinomial to the binomial theorem. We can rewrite the binomial theorem with \(k = m_1\): \[\binom{n}{m_1}=\frac{n!}{m_1!(n-m_1)!}.\] Thus, we have: \[ \begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n - m_1)!} = \binom{n}{m_1}. \] So we have shown that when \(r=2\), the multinomial reduces to the binomial.

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Most popular questions from this chapter

An urn contains \(n+m\) balls, of which \(n\) are red and \(m\) are black. They are withdrawn from the urn, one at a time and without replacement. Let \(X\) be the number of red balls removed before the first black ball is chosen. We are interested in determining \(E[X] .\) To obtain this quantity, number the red balls from 1 to \(n\). Now define the random variables \(X_{i}, i=1, \ldots, n\), by $$ X_{i}=\left\\{\begin{array}{ll} 1, & \text { if red ball } i \text { is taken before any black ball is chosen } \\ 0, & \text { otherwise } \end{array}\right. $$ (a) Express \(X\) in terms of the \(X_{i}\). (b) Find \(E[X]\).

A coin, having probability \(p\) of landing heads, is flipped until head appears for the \(r\) th time. Let \(N\) denote the number of flips required. Calculate \(E[N]\). Hint: There is an easy way of doing this. It involves writing \(N\) as the sum of \(r\) geometric random variables.

Suppose that each coupon obtained is, independent of what has been previously obtained, equally likely to be any of \(m\) different types. Find the expected number of coupons one needs to obtain in order to have at least one of each type. Hint: Let \(X\) be the number needed. It is useful to represent \(X\) by $$ X=\sum_{i=1}^{m} X_{i} $$ where each \(X_{i}\) is a geometric random variable.

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 \(X_{1}, X_{2}, \ldots, X_{n}\) be independent random variables, each having a uniform distribution over \((0,1)\). Let \(M=\operatorname{maximum}\left(X_{1}, X_{2}, \ldots, X_{n}\right) .\) Show that the distribution function of \(M, F_{M}(\cdot)\), is given by $$ F_{M}(x)=x^{n}, \quad 0 \leqslant x \leqslant 1 $$ What is the probability density function of \(M ?\)
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