## samnatha 2 years ago 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)

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1. amistre64

this might be better suited to a binomial setup

2. samnatha

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

3. amistre64

4. amistre64

the second part corresponds to an OR setup P(0) or P(1) or P(2)

5. samnatha

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

6. amistre64

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

7. samnatha

where did u get the six from ?

8. amistre64

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

9. samnatha

ok so i got 0.9728 for my final answer ?

10. samnatha

is that right ?

11. amistre64

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

12. samnatha

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 :)

13. amistre64

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