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Chapter 5: Q13E (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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