Q4.11 Consider n independent sequentia... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 4: Q4.11 (page 170)

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the n!/[k!(n 鈭� k)!] possible arrangements of the k successes and n 鈭� k failures is equally likely.

Short Answer

Expert verified

The answer is $p = \frac{p^{k} 路 (1 - p)^{n - k}}{(\begin{array}{l} n \\ k \end{array}) p^{k} (1 - p)^{n - k}} = \frac{1}{(\begin{array}{l} n \\ k \end{array})}$ .

Step by step solution

Step 1:Given Information

We have that the probability the probability for $k$ successes in $n$ trials is simply $(\begin{array}{l} n \\ k \end{array}) p^{k} (1 - p)^{n - k}$ .On the other hand ,the probability for some certain arrangement is $p^{k} 路 (1 - p)^{n - k}$ .

Step 2:Explanation

The probability for some certain arrangement given that we had $k$ successes in $n$ trials is simply $p = \frac{p^{k} 路 (1 - p)^{n - k}}{(\begin{array}{l} n \\ k \end{array}) p^{k} (1 - p)^{n - k}} = \frac{1}{(\begin{array}{l} n \\ k \end{array})}$ .

Step 3:Final Answer

The answer is $p = \frac{p^{k} 路 (1 - p)^{n - k}}{(\begin{array}{l} n \\ k \end{array}) p^{k} (1 - p)^{n - k}} = \frac{1}{(\begin{array}{l} n \\ k \end{array})}$

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91影视

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the n!/[k!(n 鈭� k)!] possible arrangements of the k successes and n 鈭� k failures is equally likely.

Short Answer

Step by step solution

Step 1:Given Information

Step 2:Explanation

Step 3:Final Answer

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