Question

help please !!
if 20% of the balls produced by the machine are defective find the probability that out of the 4 balls selected at random
(i) 2 are defective
(ii) not more than 2 are defective
i have done part (i) i got 0.1536 but i need help with part (ii)

Answer 1

this might be better suited to a binomial setup

Answer 2

thats what i did for part i and i got that i'm just not sure what to do for the other part

Answer 3

\[P(2~out~of~4)=\binom{4}{2}~(bad)^2~(good)^2\]
\[P(2~out~of~4)=6~(.20)^2~(.80)^2\]

Answer 4

the second part corresponds to an OR setup

P(0) or P(1) or P(2)

Answer 5

how do you do the second part i have done the first part ..i just need help with the second part

Answer 6

\[P(0,1,2~out~of~ 4)=6[(.2)^0 (.8)^4+(.2)^1 (.8)^3+(.2)^2 (.8)^2]\]

Answer 7

where did u get the six from ?

Answer 8

hmm, i do seem to have been a little amiss with that .... apparently we are not choosing 2 out of 4 each time :/

(4 0) = 1
(4 1) = 4
(4 2) = 6

Answer 9

ok so i got 0.9728 for my final answer ?

Answer 10

is that right ?

Answer 11

that does seem about right. can you see any other errors in my remembering of this?

Answer 12

nope although i wouldn't have added it up to 6 i would have done each sum seperatly and then added them but thank you for your help :)

Answer 13

so the liklihood that no more than 2 are defective out of 4 is about 98%, sounds like a winner to me ... good luck :)