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Chapter 5: Q 13E (page 280)

In a clinical trial with two treatment groups, the probability of success in one treatment group is 0.5, and the probability of success in the other is 0.6. Suppose that there are five patients in each group. Assume that the outcomes of all patients are independent. Calculate the probability that the first group will have at least as many successes as the second group.

Short Answer

Expert verified

The probability that the first group will have at least as many successes as the second group is 0.4957.

Step by step solution

01

Given information

The random variable X follows a binomial distribution, that is \(X \sim Bin\left( {n = 5,p} \right)\).

Two treatments are performed on two groups.

Group 1 has a probability of success that is p=0.5

Therefore, it follows, \(X \sim Bin\left( {n = 5,p = 0.5} \right)\)

Group 2 has a probability of success that is p=0.6

Therefore, it follows, \(X \sim Bin\left( {n = 5,p = 0.6} \right)\)

02

Formulate binomial probabilities table

03

Calculating the probability

The probability that the first group will have at least as many successes as the second group is:

\(0.03120{\rm{ }} \times 0.0102{\rm{ }} + {\rm{ }}0.1562{\rm{ }} \times {\rm{ }}\left( {0.0102{\rm{ }} + {\rm{ }}0.0768} \right){\rm{ }} + {\rm{ }}...{\rm{ }} + {\rm{ }}0.0312{\rm{ }} \times {\rm{ }}1.\)

The final answer is 0.4957.

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Most popular questions from this chapter

Suppose that X1 and X2 form a random sample of twoobserved values from the exponential distribution with parameter \({\bf{\beta }}\). Show that \(\frac{{{X_1}}}{{\left( {{X_1} + {X_2}} \right)}}\) has the uniform distribution on the interval [0, 1].

Let F be a continuous CDF satisfying F (0) = 0, and suppose that the distribution with CDF F has the memoryless property (5.7.18). Define f(x) = log[1 − F (x)] for x > 0

a. Show that for all t, h > 0,

\({\bf{1 - F}}\left( {\bf{h}} \right){\bf{ = }}\frac{{{\bf{1 - F}}\left( {{\bf{t + h}}} \right)}}{{{\bf{1 - F}}\left( {\bf{t}} \right)}}\)

b. Prove that \(\ell \)(t + h) = \(\ell \)(t) + \(\ell \)(h) for all t, h > 0.

c. Prove that for all t > 0 and all positive integers k and m, \(\ell \)(kt/m) = (k/m)\(\ell \) (t).

d. Prove that for all t, c > 0, \(\ell \) (ct) = c\(\ell \) (t).

e. Prove that g(t) = \(\ell \) (t)/t is constant for t > 0

f. Prove that F must be the CDF of an exponential distribution.

Let \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) denote the p.d.f. of the bivariate normaldistribution specified by Eq. (5.10.2). Show that the maximumvalue of \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) is attained at the point at which \({{\bf{x}}_{\bf{1}}} = {{\bf{\mu }}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{x}}_{\bf{2}}} = {{\bf{\mu }}_{\bf{2}}}.\)

Suppose that the measured voltage in a certain electric circuit has the normal distribution with mean 120 and standard deviation 2. If three independent measurements of the voltage are made, what is the probability that all three measurements will lie between 116 and 118?

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\) are independent and that \({{\bf{X}}_{\bf{i}}}\) has the Poisson distribution with mean \({{\bf{\lambda }}_{\bf{i}}}\left( {{\bf{i = 1, \ldots ,k}}} \right)\). Show that for each fixed positive integer n, the conditional distribution of the random Vector \({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}} \right)\), given that \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{{\bf{X}}_{\bf{i}}}{\bf{ = n}}} \) it is the multinomial distribution with parameters n and

\(\begin{array}{l}{\bf{p = }}\left( {{{\bf{p}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{p}}_{\bf{k}}}} \right){\bf{,}}\,{\bf{where}}\\{{\bf{p}}_{\bf{i}}}{\bf{ = }}\frac{{{{\bf{\lambda }}_{\bf{i}}}}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{k}} {{{\bf{\lambda }}_{\bf{j}}}} }}\,{\bf{for}}\,{\bf{i = 1, \ldots ,k}}{\bf{.}}\end{array}\)
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