# A question about card probability

#### FillenNaymeer

##### New member
If I had a deck of cards with 40 blue cards and 58 red cards, so a total of 98 cards shuffled randomly, and i drew a hand of 7 cards, what is the probability that I draw 3 or more blue cards in the hand?
Thanks for any help you all can offer, a solution with some explanation would be greatly appreciated.

#### Dr.Peterson

##### Elite Member
If I had a deck of cards with 40 blue cards and 58 red cards, so a total of 98 cards shuffled randomly, and i drew a hand of 7 cards, what is the probability that I draw 3 or more blue cards in the hand?
Thanks for any help you all can offer, a solution with some explanation would be greatly appreciated.
I would start by finding the probability of drawing fewer than 3 blue cards (that is, 0, 1, or 2). That gives you only three calculations to do.

Now please show some work, so we can see where you need help.

#### FillenNaymeer

##### New member

I would start by finding the probability of drawing fewer than 3 blue cards (that is, 0, 1, or 2). That gives you only three calculations to do.

Now please show some work, so we can see where you need help.
Thanks for the quick reply. There was a typo in the answer key that had me thinking the pmf method wasn't right so I started overthinking it and confusing myself. I'm kinda still wondering why we can use a constant probability of success and failure for all the instances of blue cards and red cards being drawn in the hand of 7. It feels like the probability should change as the cards are drawn going from 40/98 to 39/97 for example for the probability of drawing a blue card after a blue card is already drawn since there's no replacement. I know that's not how it works but I don't really understand why that's not how it works I guess. Thanks again for the help

#### Dr.Peterson

##### Elite Member
View attachment 26320

Thanks for the quick reply. There was a typo in the answer key that had me thinking the pmf method wasn't right so I started overthinking it and confusing myself. I'm kinda still wondering why we can use a constant probability of success and failure for all the instances of blue cards and red cards being drawn in the hand of 7. It feels like the probability should change as the cards are drawn going from 40/98 to 39/97 for example for the probability of drawing a blue card after a blue card is already drawn since there's no replacement. I know that's not how it works but I don't really understand why that's not how it works I guess. Thanks again for the help
The probability would change! The binomial distribution is not appropriate for this problem, because cards are drawn without replacement. What makes you think it is? Why would you let that convince you that you don't understand reality?

Are you saying that the answer key gave the answer you got?

(I wish people would show all the information they have, including book answers, so that we could more quickly see what the real issue is.)

#### Jomo

##### Elite Member
1-.402 is NOT .597