Thread: approximations of distributions 2

A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

I find:
p = 1/6



But the answer is: 0,4129

Which side of 17,5 do you want? 1 - 0,587064 = 0,412936 That looks pretty close.

Originally Posted by Apprentice123

A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

I find:
p = 1/6



But the answer is: 0,4129

You have calculated the probability of fewer than 18 not 18 or more.

CB

Not is P(X >= x) = (P X >= x-0,5) ???

Originally Posted by Apprentice123

Not is P(X >= x) = (P X >= x-0,5) ???

What does that mean? (I'm betting whatever you intend it to mean the answer is no)



CB