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# Random Variables (Binomial) watch

1. Hello Just a question on Binomial Distribution:

THe Random variable X has the distribution B(n,0.4)

Give that n = 10, find the smallest value of x such that P(X>x) <0.05

(b) Find the smallest value of n such that P(X=0) isless than 0.001

Thanks
From John
2. (Original post by Neo1)
Hello Just a question on Binomial Distribution:

THe Random variable X has the distribution B(n,0.4)

Give that n = 10, find the smallest value of x such that P(X>x) <0.05

(b) Find the smallest value of n such that P(X=0) isless than 0.001

Thanks
From John
P(X>x) = 1-P(X =< x)
1-P(X =< x) < 0.05
P(X =< x) > 0.95
use table for n=10 from here onwards (with p=0.4)

b) P(X=0) < 0.001
nC0 * 0.6^n < 0.001
0.6^n < 0.001
n = log(0.001)/ log(0.6) (rounded up)
3. I didnt get part (b) very well do you mind explaining please and I havent done logs yet so what other method could I use

Thanks a lot
From John
4. b) P(X=0) < 0.001

binomial formula is [for X~B(n,p) ]:
P(X=r) = n!/(r!(n-r)! * p^r * (1-p)^(n-r)

n!/0!n! * 0.6^n < 0.001

0.6^n < 0.001
n = log(0.001)/ log(0.6) (rounded up)

there's no "proper" way to do it apart from logs. having said that you could type ANS = 0.6 into your calculator. then type ANS*0.6 and see how many time you have to press "equals" before the display shows a number less than 0.001.

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