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Mathematics 19 Online
OpenStudy (anonymous):

PLEASE HELP fan and medal... 2% of skiers get injured going down a certain hill. 1.What is the probability that exactly 3 of the hundred skiers will injure themselves?

OpenStudy (tkhunny):

p(exactly 0) = .98^100 p(exactly 1) = 100 * .98^99 * .02 p(exactly 2) = 100*99/2 * .98^98 * .02^2 p(exactly3) = ?? What say you?

OpenStudy (anonymous):

.126?

OpenStudy (anonymous):

@tkhunny would it be .126?

OpenStudy (tkhunny):

You tell me. How did you get that?

OpenStudy (anonymous):

@tkhunny well i was a little confused, but i skipped to the third equation and plugged that into my calculator

OpenStudy (tkhunny):

p(0) = 0.133 p(1) = 0.271 p(2) = 0.273 p(3) = 0.182 = \(\dfrac{100*99*98}{3*2}\cdot0.98^{97}\cdot 0.02^3\) Looks like you're a little off. You will need to know this rule for your exam. Sometimes, it's convenient just to build the entire distribution. Other times, like this one, don't do that.

OpenStudy (anonymous):

.182, thanks! @tkhunny

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