A new drug that treats patients with anxiety has a 60% success rate in improving the patient’s condition.
For a random sample of eight patients taking this drug, determine the probability that:
4.2.1. four patients have improvement in their condition.
4.2.2. at least two patients have improvement in their condition
To determine the probability of certain outcomes, we can use the binomial probability formula. The formula for the probability of obtaining exactly x successes in n trials, given a success rate of p, is:
P(X = x) = (nCx) * p^x * (1-p)^(n-x)
where nCx represents the number of combinations of n items taken x at a time.
Let's calculate the probabilities for the given scenarios:
4.2.1. Probability of four patients having improvement in their condition:
In this case, we have a random sample of eight patients, and we want to find the probability that exactly four patients have improvement. The probability of success (p) is given as 0.6.
P(X = 4) = (8C4) * (0.6^4) * (1-0.6)^(8-4)
Using the combination formula (nCx = n! / (x!(n-x)!)), we have:
P(X = 4) = (8! / (4!(8-4)!)) * (0.6^4) * (0.4^4)
= (8! / (4!4!)) * (0.6^4) * (0.4^4)
Calculating this expression, we find the probability that exactly four patients have improvement in their condition.
4.2.2. Probability of at least two patients having improvement in their condition:
In this case, we want to find the probability that at least two patients have improvement in their condition. This includes scenarios where two, three, four, five, six, seven, or eight patients have improvement.
P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 7) + P(X = 8)
Calculate each of the probabilities using the formula shown above for each value of X and sum them up.
P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Now, calculate this expression to find the probability that at least two patients have improvement in their condition.
By following these steps and calculating the appropriate probabilities, you can determine the desired likelihoods based on the success rate and sample size given.