SOLUTION: I need help on this problem There is about an 80% chance that a 20 year old today will be alive at the age of 65. Suppose five 20 year olds were selected at random. What are the

Algebra ->  Probability-and-statistics -> SOLUTION: I need help on this problem There is about an 80% chance that a 20 year old today will be alive at the age of 65. Suppose five 20 year olds were selected at random. What are the      Log On


   



Question 536397: I need help on this problem
There is about an 80% chance that a 20 year old today will be alive at the age of 65. Suppose five 20 year olds were selected at random. What are the following probabilities?
a) P(exactly 3 will be alive at 65)
b) P(at least 3 will be alive at 65)
c) P(at most 2 will be alive at 65)
Thank you

Answer by fcabanski(1391) About Me  (Show Source):
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Probability of an event, P, occurring exactly r times is:


nCp%2Ap%5Er%2Aq%5E%28n-r%29 where n is the number of trials (5 in this case), p is the probability (.8), r the number desired (such as 3 students being alive), and q=1-p which is .2 in this case.


nCr=n%21%2Fr%21%28n-r%29%21


Often we solve such problems by adding a number of probabilities. For example, if it's the probability that "at least" 2 survive, we add the probability for 2, 3, 4 and 5 found with the above formula.


1. P(exactly 3 alive) gives r =3. So P = = 20.48%


2. P(at least 3 alive) means r%3E=3 so add the P(3) + P(4) + P(5).


P(3)= 20.48%


P(4)= = 40.96%


P(5) = = 32.77%


P(3 or more) = P(3)+P(4)+P(5)=20.48+40.96+32.77 = 94.21%


3. P(at most 2) means r%3C=2 so add P(2) + P(1) + P(0).


P(2)= = 5.12%


P(1) = = .64%


P(0) = = .032%


P(at most 2) = P(2)+P(1)+P(0)= 5.12% + .64% +.032% = 5.79%