7. Bud and his friends stopped at a local fast food restaurant after the football game Friday night. Bud ate a one-fourth pound
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Answer:
0.0923 = 9.23% probability that four of the patients are still alive after five years.
Step-by-step explanation:
For each patient, there are only two possible outcomes. Either they are still alive after five years, or they are not. The probability of a patient being alive is independent of any other patient, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Po is an oncologist with seven patients in their care.
This means that
The probability that a patient will survive five years after being diagnosed with stage three breast cancer is 0.82.
This means that
What is the probability that four of the patients are still alive after five years?
This is P(X = 4). So
0.0923 = 9.23% probability that four of the patients are still alive after five years.