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 can start by finding the probability of each possible outcome for the number of successes.
Let's break down the problem into steps:
Step 1: Define the probabilities of success for each drug.
Given that p1 = p2 = p3 = 0.7, we know that each drug has a 70% chance of success.
Step 2: Determine the probability of each possible outcome.
Since there are 10 "triples," each with three possible drugs, there are a total of 3^10 = 59049 possible outcomes. To compute the probability of each outcome, we need to consider the number of successes for each drug.
- To have exactly 0 successes with one drug, we need to distribute the 10 triples among the three drugs. This can be calculated using the multinomial coefficient formula: (10 choose 0,0,10) = 1.
- To have exactly 1 success with one drug, we need to distribute 9 triples among two drugs and 1 triple for the other drug. Again, using the multinomial coefficient formula: (10 choose 1,1,8) = 45.
- To have exactly 2 successes with one drug: (10 choose 2,2,6) = 252.
- To have exactly 3 successes with one drug: (10 choose 3,3,4) = 840.
- To have exactly 4 successes with one drug: (10 choose 4,4,2) = 1260.
- To have exactly 5 successes with one drug: (10 choose 5,5,0) = 252.
Step 3: Compute the probability that the maximum number of successes exceeds eight.
To compute this probability, we need to sum the probabilities of all outcomes where the maximum number of successes is greater than eight. In this case, it means considering the probabilities of having 4, 5 successes with one drug.
Probability = (probability of 4 successes) + (probability of 5 successes)
Probability = (1260/59049) + (252/59049)
Probability = 1512/59049
Therefore, the probability that the maximum number of successes with one of the drugs exceeds eight, given the conditions above, is approximately 0.0256 (or 2.56%).