I need a doctor for this question~!
If 2% of the batteries manufactured by a company are defective, find the probability that in case of 144 batteries, there are 3 defective ones. Tq ^^

Question

amistre64 · Answer

p = .02
q = .98

has something to do with those I think

anonymous · Answer

is it by using binomial distribution?

amistre64 · Answer

i cant recall :)

amistre64 · Answer

prolly

jamesj · Answer

Yes.  Now what is P(none defective) ?  = q^144
P(1 defective) = nC1 q^143.p
P(2 defective) = nC2 q^142.p^2
P(3 defective) = nC3 q^141.p^3

These are also the first four terms of the binomial expansion of
(p+q)^144

amistre64 · Answer

$${144\choose 3}p^{.02}q^{.98}$$
maybe?

amistre64 · Answer

got my ps and qs misappropriated lol

jamesj · Answer

(  I should have written 144 for n above )

anonymous · Answer

$$\dbinom{144}{3}(.02^3)(.98)^{141}$$ for 3 defective, others are similar

anonymous · Answer

thats is for each one use 
$$\dbinom{n}{k}p^k(1-p)^{n-k}$$ with 
$$k=0,1,2,3, p=.02,1-p=.98$$

anonymous · Answer

oh right, what jamesj said above!

anonymous · Answer

would help if i read the question.. it says "3 are defective"

anonymous · Answer

$$\dbinom{144}{3}(.02)^3(.98)^{141}$$ is the one number you need

anonymous · Answer

tq! ans : 225.84