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in testing a new drug. researchers found that 20% of all patients using it will have a mild side effect. a random sample of 14...Asked by Jen
3. In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that:
(A) exactly two will have this mild side effect
(B) at least three will have this mild side effect.
I got this for (A):
= 14!/(14-2)!2! (0.10)²(0.90)14-2^2
= 14*13*12!/12!2*1(0.10)²(0.90)12^2
= 14*13/2*1(0.01)(0.28)
= 182/2(0.01)(0.28)
= 91(0.01)(0.28) = 0.25 or 25%
But i'm not sure how to do (B)
PLZZZZZ HELP!!!
(A) exactly two will have this mild side effect
(B) at least three will have this mild side effect.
I got this for (A):
= 14!/(14-2)!2! (0.10)²(0.90)14-2^2
= 14*13*12!/12!2*1(0.10)²(0.90)12^2
= 14*13/2*1(0.01)(0.28)
= 182/2(0.01)(0.28)
= 91(0.01)(0.28) = 0.25 or 25%
But i'm not sure how to do (B)
PLZZZZZ HELP!!!
Answers
Answered by
MathGuru
A looks OK!
For B, use the same process. Since the problem says "at least 3" you will have to determine P(0), P(1), P(2). You already have P(2) from part A. Add P(0), P(1), and P(2) together. Then subtract that value from 1 for your probability.
I hope this helps.
For B, use the same process. Since the problem says "at least 3" you will have to determine P(0), P(1), P(2). You already have P(2) from part A. Add P(0), P(1), and P(2) together. Then subtract that value from 1 for your probability.
I hope this helps.
Answered by
spam
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