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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?

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    Thanks Now I get It , You are best teacher !2012-11-18

1 Answers 1

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(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= $1-\frac{\binom {37}5}{\binom {40}5}$

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    @Hooman, welcome. This answer clearly assumes no replacement.2012-11-18