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September 1, 2014

September 1, 2014

Posted by **Jen** on Wednesday, October 26, 2011 at 6:19pm.

(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!!!

- Did i do this right??? STATS -
**MathGuru**, Thursday, October 27, 2011 at 5:28pmA 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.

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