2 Binomial Probability Questions

xibeleli · #1 $Old$ June 7th, 2009, 07:40 AM

The probability of an integrated circuit passing a quality test is 0.86. In a batch of 10 integrated circuits, find the probability that

iii) at least 2 will fail
iv) fewer than 2 will fail

I know the forumula to use

P(x) = NCr (a)^r (b)^r-1

but when it gets "least" and "fewer" i start to get confused
whats the easy to this i know its 1 - p(something) but not really sure

Help please guys

Plato · #2 $Old$ June 7th, 2009, 08:04 AM

Quote:

Originally Posted by xibeleli $View Post$

The probability of an integrated circuit passing a quality test is 0.86. In a batch of 10 integrated circuits, find the probability that
iii) at least 2 will fail
iv) fewer than 2 will fail

The probability that none or one fails is $\sum\limits_{k = 0}^1 {\binom{10}{k}\left( {.86} \right)^{10 - k} \left( {.14} \right)^k }$ .
That answers (iv).

matheagle · #3 $Old$ June 7th, 2009, 08:36 AM

At least 2 means X is 2 or 3 or 4 or... 10. Here you want $P(X\ge 2)$ .

Fewer than 2 means X is 0 or 1. Here you want $P(X < 2)=P(X\le 1)$ .

Note that $P(X\ge 2)+P(X\le 1)=1$ .

And it's n-r, not r-1.