1. ## Probability

if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random,

a 1 defective

b none defective

c at most 2 bolts will be defective

think i have it just not sure?

2. Originally Posted by question
if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random,

a 1 defective

b none defective

c at most 2 bolts will be defective

think i have it just not sure?
A) $P(1) = .2^1(.8)^3$ $4\choose1$ $\Rightarrow .4096$

3. this is a binomial distribution problem. see if post #2 here helps

4. Originally Posted by colby2152
A) $P(1) = .2^1(.8)^3$ $4\choose1$ $\Rightarrow .4096$

could u possibly post the formula in its pure form with its components explained

5. Originally Posted by question
could u possibly post the formula in its pure form with its components explained
i did that in the post i directed you to

6. Originally Posted by question
if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random,

a 1 defective

b none defective

c at most 2 bolts will be defective

think i have it just not sure?
for part c i have the following:

(4c2).(0.2)^2.(0.8^2) = 0.1536

this is the answer that we were supposed to arrive at, but just looking at the wording of the question im not sure if its correct.

Can anyone verify???

7. Originally Posted by question
for part c i have the following:

(4c2).(0.2)^2.(0.8^2) = 0.1536

this is the answer that we were supposed to arrive at, but just looking at the wording of the question im not sure if its correct.

Can anyone verify???
no. that is if EXACTLY two are defective. "at most two" means 2 or less. meaning 0 or 1 or 2 (or more easily, not(3 or 4))

8. Originally Posted by question
if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random,

a 1 defective

b none defective

c at most 2 bolts will be defective

think i have it just not sure?
BTW, the solution to c will start off like this:

P(at most 2) $= P(0) + P(1) + P(2) \Rightarrow 1 - P(3) - P(4)$