Three drugs are being tested for use as the treatment of a certain disease.
Let p 1 , p 2 , and p 3 represent the probabilities of success for the respective drugs. As three patients come in, each is given one of the drugs in a random order. After n = 10 “triples” and assuming independence, compute the probability that the maximum
number of successes with one of the drugs exceeds eight if in fact p 1 = p 2=p 3 = 0.7
To compute the probability that the maximum number of successes with one of the drugs exceeds eight, we need to calculate the probability of each possible outcome and then sum up the probabilities of the desired outcomes.
First, let's determine the probability of success for each drug. Given that p1 = p2 = p3 = 0.7, the probability of success for each drug is 0.7.
Next, let's calculate the probability of each outcome for a single triple. We have three drugs, and each patient is given one of the drugs in a random order. We can represent each outcome as a combination of the number of successes for each drug.
Let's consider the possible outcomes:
- 3 successes (all patients receive the drug with a probability of success 0.7)
- 2 successes (two patients receive the drug with a probability of success 0.7, and the remaining one receives a different drug)
- 1 success (one patient receives the drug with a probability of success 0.7, and the other two receive different drugs)
- 0 successes (all patients receive different drugs)
Now, let's calculate the probability for each outcome:
Probability of 3 successes: (0.7 * 0.7 * 0.7) = 0.343
Probability of 2 successes:
- Two patients receive the drug with a probability of success 0.7: (0.7 * 0.7 * 0.3) = 0.147
- The remaining patient receives a different drug with a probability of success 0.7: (0.7 * 0.3 * 0.7) = 0.147
- Multiply the above two probabilities by 3 to account for different arrangements: (0.147 + 0.147) * 3 = 0.882
Probability of 1 success:
- One patient receives the drug with a probability of success 0.7: (0.7 * 0.3 * 0.3) = 0.063
- The other two patients receive different drugs with a probability of success 0.7: (0.3 * 0.7 * 0.3) + (0.3 * 0.3 * 0.7) = 0.189
- Multiply the above two probabilities by 3 to account for different arrangements: (0.063 + 0.189) * 3 = 0.816
Probability of 0 successes: (0.3 * 0.3 * 0.3) = 0.027
Now, we need to calculate the probability that the maximum number of successes exceeds eight out of a total of 10 triples.
To do this, we will calculate the probability of each possible number of triples where the maximum number of successes is greater than eight and then sum them up.
Possible scenarios:
- 9 triples with 9 successes: Probability = (0.343) ^ 9 = 0.0004478
- 10 triples with 10 successes: Probability = (0.343) ^ 10 = 0.0001536
Summing these probabilities up: 0.0004478 + 0.0001536 = 0.0006014
Therefore, the probability that the maximum number of successes with one of the drugs exceeds eight is approximately 0.0006014 when p1 = p2 = p3 = 0.7.