Thread: last one....

There are 100 cards in a set. I have 98 of them. The cards are sold in packs of 7. I need 2 more to complete the set.

If i buy one pack what is the probability that the first card is one that i need?

I got 1/50.

Find the probability that i find neither of the cards i need in the 7 cards of the pack.

I got 43/50.

Let the event that I get a card that i need = C and the event that i dont Cdash. FInd the probability that i get exactly one card that i need. (edit: in the 7 cards if one pack)

I am stuck...
Thanks!

You need to give more information about the contents of the set of 7. Are they (i) seven randomly chosen cards, or (ii) seven different cards, but otherwise random?

In case (i) or (ii) the first card is a randomly chosen card out of 100 and the chance that it is one of the two you want is 2/100 = 1/50.

In case (i) the chance that each of the seven cards is not one you want is 98/100 and so the chance that all seven are not ones you want is (98/100)^7. In case (ii) the chance that the seven are drawn from the 98 you don't want as opposed to being drawn from all 100 possible is 98-choose-7 / 100-choose-7 = (98!/91!.7!)/(100!/93!.7!) = 98!.93!.7! / 100!.91!.7! which fortunately cancels down to 93.92/100.99.

The set of 7 are 7 random cards which all are different.