# Thread: Probability Q12

1. ## Probability Q12 [ NEED HELP FOR PART C ]

Attempt for a and b are fine. However c is wrong!!! require help for part C

There are three cartons of vases, each containing 8 good vases and 2 defective ones. A buyer selects 2 vases at random from each carton to check. If both vases selected from a carton are good, he will accept the carton: otherwise, he will reject the carton. Find the probability that

(a) he will accept a carton.
(b) he will accept all three cartons.
(c) he will accept at least one carton.

Attempt

a) P(he will accept a carton)= $[\frac{8}{10}][\frac{7}{9}$
= $\frac{28}{45}$

b)P(he will accept all three)= $[\frac{28}{45}]^3$
= $\frac{21952}{91125}$

c)P(at least 1 accepted)= $1-P(he will not accept any)$
= $1-(\frac{2}{10}\times\frac{1}{9})^3$
= $0.999$

2. P( he will not accept any ) = 1 - ( 2/90 )^3 = 0.999

did you forgot to write that that or? ( than it is ok )

3. Sorry, that was just a typo. In actual fact, I have already take the power of 3 into account.

4. Does anybody have different thoughts/workings from me? Please post them!

5. Originally Posted by Punch
Does anybody have different thoughts/workings from me? Please post them!
For (c), note that Pr(reject carton) = 1 - Pr(accept carton) = 1 - 56/90 = 34/90 ....