4.2. 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 calculate the probability in both scenarios, we will be using the binomial probability formula.
The binomial probability formula is as follows:
P(X = k) = nCk * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k successes
- nCk is the number of ways to choose k items from a set of n (also known as combinations)
- p is the probability of success for a single trial
- q = (1-p) is the probability of failure for a single trial
- k is the number of successes
- n is the total number of trials
Now let's proceed to calculate the probabilities:
4.2.1. Four patients have improvement in their condition:
To calculate the probability that exactly four patients have improvement in their condition, we use the binomial probability formula.
nCk = 8C4 = 8! / (4! * (8-4)!) = 70
p^k = 0.6^4 = 0.1296
(1-p)^(n-k) = (1-0.6)^(8-4) = 0.1296
P(X = 4) = 70 * 0.1296 * 0.1296 ≈ 0.7129
Therefore, the probability that four patients have improvement in their condition is approximately 0.7129.
4.2.2. At least two patients have improvement in their condition:
To calculate the probability that at least two patients have improvement in their condition, we sum up the probabilities of getting exactly two, three, four, five, six, seven, and eight successes.
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Using the binomial probability formula for each case:
P(X = 2) = 8C2 * 0.6^2 * 0.4^6 ≈ 0.1209
P(X = 3) = 8C3 * 0.6^3 * 0.4^5 ≈ 0.2852
P(X = 4) ≈ 0.7129 (calculated previously)
P(X = 5) = 8C5 * 0.6^5 * 0.4^3 ≈ 0.1935
P(X = 6) = 8C6 * 0.6^6 * 0.4^2 ≈ 0.0439
P(X = 7) = 8C7 * 0.6^7 * 0.4^1 ≈ 0.0055
P(X = 8) = 8C8 * 0.6^8 * 0.4^0 ≈ 0.0002
P(X ≥ 2) = 0.1209 + 0.2852 + 0.7129 + 0.1935 + 0.0439 + 0.0055 + 0.0002 ≈ 1
Therefore, the probability that at least two patients have improvement in their condition is approximately 1 (or 100%).