# Probability of 1 and 2 are not among 5 cards chosen from 40

Suppose we have 40 cards numbered 1 through 40 and we want to pick 5 of them the at the same time.

a) Probability of 1 and 2 are not among those 5 cards

b) Probability of at least one of 10,11,12 are among the five

* How about picking cards with replacemnt?

-
If with replacement, probability first card is not $1$ or $2$ is $38/40$. Probability this happens $5$ times in a row is $(38/40)^5$. Probability $10$, $11$, and $12$ are all missing on one pick us $(37/40)$. Probability of this $5$ times in a row is $(37/40)^5$. So probability $11$, $12$, or $13$ occurs at least once is $1-(37/40)^5$. –  André Nicolas Nov 18 '12 at 16:36
One last question ,if we do this game but this time "without replacemnt" are we going to get the same result as "at the same time" in both parts a and b? –  Hooman Nov 18 '12 at 16:40
How about the sample space in "with replacement", if we consider sample space to be 40^5 I think we have considered the order of picking cards so we need to divide it by 5! , am I right ? –  Hooman Nov 18 '12 at 16:47
Just to make sure I understand completely, 40^5 gives us the sample space with order does't matter and (40^5)*5! gives us sample space with order matters ? –  Hooman Nov 18 '12 at 16:58
I am sorry but Right now I am completely confused, also Thanks a lot for help and quick responding, In this problem I think it is asking for order doesn’t matter when we pick the cards so when we say 40^5 actually order does matter and when we say C(40,5) order doesn’t matter So what is the difference in sample space when we want with replacement and without replacement ? (This type of questions are always confusing me), Thanks Man –  Hooman Nov 18 '12 at 17:16

(a)If we don't allow $1,2;$ we have to choose $5$ cards from the rest $(40-2)=38$ cards. SO, the required probability = the number favourable case/all the case = $$\frac{\binom {38}5}{\binom {40}5}$$

(b) Similarly, if we don't allow $10,11,12;$ we are left with $40-3=37$ cards.

So, the probability of choosing $5$ cards without $10,11,12$ is $$\frac{\binom {37}5}{\binom {40}5}$$

Hence, the required probability= $$1-\frac{\binom {37}5}{\binom {40}5}$$

-
is't it the same answer as we pick the card with out replacement ? –  Hooman Nov 18 '12 at 16:27
How about picking cards with replacemnt? –  Hooman Nov 18 '12 at 16:33
@Hooman, what do you exactly mean by replacement? –  lab bhattacharjee Nov 18 '12 at 16:35
after picking card we put it back and then picking anther card and so on –  Hooman Nov 18 '12 at 16:36
@Hooman, please find the comment of "André Nicolas " in the question. –  lab bhattacharjee Nov 18 '12 at 16:37