# Thread: probability of missing items

1. ## probability of missing items

i have problem of finding P(C) , why cant i use the method as in the photo to find P(C) beacuse 1-(all unused stamps are missing)= at least 1 use stamps is missing.

2. ## Re: probability of missing items

You can do that but you have incorrectly calculated the probability that all of the missing stamps are unused.

You have 24 stamps, 10 of which are 'unused'. The probability the first stamp missing is unused is 10/24= 5/12. There are then 23 stamps, 9 of which are unused. The probability the second stamp missing is unused is 9/23. There are then 22 stamps, 8 of which are unused. The probability the third stamp missing is unused is 8/22= 4/11. The probability all three missing stamps are unused is $\displaystyle \frac{5}{12}\frac{9}{23}\frac{4}{11}= \frac{6}{101}$ so the probability there is at least one used stamp among the missing stamps is $\displaystyle 1- \frac{6}{101}= \frac{95}{101}$.

3. ## Re: probability of missing items

Why the unused not 2 +6+2+4 = 14?

4. ## Re: probability of missing items

Originally Posted by delso
Why the unused not 2 +6+2+4 = 14?
I don't know what you mean that. The total number of "unused stamps" is 14 but that is not directly relevant to the problem.

5. ## Re: probability of missing items

$\displaystyle P(C) = 1-\dfrac{14\cdot 13\cdot 12}{24\cdot 23\cdot 22} = \dfrac{415}{506}$.

$\displaystyle P(C|A) = 1-\dfrac{10\cdot 9\cdot 8}{20\cdot 19\cdot 18} = \dfrac{17}{19}$

If no green stamps are missing, then there are only 10 possible unused stamps that can be missing. So, the probability that at least one stamp is used is the complement of the probability that all three stamps are unused (none green). Essentially, you pretend that green stamps don't exist.

6. ## Re: probability of missing items

i dont have ans for this question. is my ans for the previous part correct?