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A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 4: Q.4.15 (page 170)

Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is
$\frac{1}{2} [1 + (q - p)^{n}]$ .where $q = 1 - p$ .Do this by proving and then utilizing the identity
${鈭�}_{i = 0}^{[n / 2]} (\begin{matrix} n \\ 2 i \end{matrix}) p^{2 i} q^{n - 2 i} = \frac{1}{2} [(p + q)^{n} + (q - p)^{n}]$

Short Answer

Expert verified

In the given information the answer is $= \frac{1}{2} [1 + (q - p)^{n}]$ isproved

Step by step solution

01

Given Information

We know that ,

${(p + q)}^{n}$ $= {鈭�}_{i = 0}^{n} {}^{n}C_{k} p^{k} q^{n + 1}$ (1)

${(q - p)}_{n}$ $(q - p)^{n} = {鈭�}_{i = 0}^{n} {}^{n}C_{k} (- p)^{k} q^{n - k}$ (2)

02

Calculation

When we add these two expressions ,

$鈬� {(p + q)}^{n} + {(q - p)}^{n} =$ $2 {鈭�}_{i = 0}^{\frac{n}{2}} C_{2 i} p^{2 i} q^{n - 2 i}$

because when $k$ is odd ${(- p)}^{k}$ is negative so (1) and (2) will cancel .

$鈬� {鈭�}_{i = 0}^{\frac{n}{2}} C_{2 i} p^{2 i} q^{n - 2 i} = \frac{1}{2} [(p + q)^{n} + (q - p)^{n}]$

$鈬� p (E v e n h e a d s)$ = $\frac{1}{2} [(p + 1 - p)^{n} + (q - p)^{n}]$

$= \frac{1}{2} [1 + (q - p)^{n}]$ hence proved

03

Final Answer

The answer is $= \frac{1}{2} [1 + (q - p)^{n}]$ is proved $\frac{1}{2} [1 + (q - p)^{n}]$

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Most popular questions from this chapter

Consider n independent trials, each of which results in one of the outcomes $1, . . ., k$ with respective probabilities $p_{1}, 鈥�, p_{k}, {鈭�}_{i = 1}^{k} p_{i} = 1$ Show that if all the $p i$ are small, then the probability that no trial outcome occurs more than once is approximately equal to $\exp (- n (n - 1) \underset{i}{鈭�} p_{i}^{2} / 2)$ .

Suppose that $X$ takes on one of the values $0, 1$ and $2$ . If for some constant $c, P {X = i} = c P {X = i - 1}, i = 1, 2$ , find $E [X] .$

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A student is getting ready to take an important oral examination and is concerned about the possibility of having an 鈥渙n鈥� day or an 鈥渙ff鈥� day. He figures that if he has an on the day, then each of his examiners will pass him, independently of one another, with probability $8$ , whereas if he has an off day, this probability will be reduced to $4$ . Suppose that the student will pass the examination if a majority of the examiners pass him. If the student believes that he is twice as likely to have an off day as he is to have an on the day, should he request an examination with $3$ examiners or with $5$ examiners?

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