Card probability

[lordanyc]

Hello all - I was wondering if you could help me with equation how to solve this, would very much appreciate it!

There are 2 decks of cards:
- deck #1 has 100 statement cards
- deck #2 has 400 answer cards
- each statement card has 40 possible matching cards (GOOD MATCH)

Question
1. At random I pick 10 answer cards from pile of 400
2. I then at random pick 1 statement card from pile of 100
3. What is the probability I will end up with a GOOD MATCH answer to my statement?

Thanks very much!

 

[pka]

The one card you pick pile #1 corresponds to forty good cards in pile #2. Go among the ten cards your pick from pile #2 you want at least one matching good card. Well in pile #2 there are three hundred and sixty cards that do not match. If I were you I would calculate the probability of not getting any matching cards among the ten. There are \(\mathcal{C}_{10}^{400}\) ways to draw ten cards from pile #2. \(\mathcal{C}_{10}^{360}\) ways to draw ten cards from pile #2 that do not match the the one from pile #1.
So \(1-\left(\dfrac{\mathcal{C}_{10}^{360}}{\mathcal{C}_{10}^{400}}\right)\)

 

[lordanyc]

Thanks Pka! Apologies for my dumb question but what does C represent, is that just 10/400? Also your equation has numerator and denominator the same? Is that correct?

 

[pka]

what does C represent, is that just 10/400? Also your equation has numerator and denominator the same? Is that correct?

Combinations \(\mathcal{C}_{k}^{N}=\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}\) that is a combination of \(N\) distinct objects taken \(k\) at a time.
In what I posted \(\mathcal{C}_{10}^{400}=\dfrac{400!}{(10!)(390!)}\). Learn to use this site.

 

[lordanyc]

thanks but its still not clear to me your initial posting had numerator and denominator same. is this correct? I appreciate your help, but probably many people are not as experienced in this and this is quite foreign so if I look at below top and bottom are same


That was a typo - it has been fixed. p\Post #2 should have been

\(\displaystyle 1-\left(\dfrac{\mathcal{C}_{10}^{360}}{\mathcal{C}_{10}^{400}}\right)\)

 

[Steven G]

thanks but its still not clear to me your initial posting had numerator and denominator same. is this correct? I appreciate your help, but probably many people are not as experienced in this and this is quite foreign so if I look at below top and bottom are same
View attachment 18974

Since the numerator and denominator are the same, and the by the definition given to you by pka both the numerator and denominators are real numbers, then that fraction is 1 and 1-1=0. Do you think that the answer to your question is 0?

 

[pka]

EDIT

IT should be So \(\huge{1-\left(\dfrac{\mathcal{C}_{10}^{360}}{\mathcal{C}_{10}^{400}}\right)}\)

 

[lordanyc]

Thanks Pka.. much appreciate your help here.. so if I understand correctly if I used your formula

400 choose 10 = 25798075602615500000
360 choose 10 = 8881600973913290000

so then if I do calculation I get answer .66 = 66% so if understand your logic correctly is this probability of finding a matching card? (I assume this is a reason you added 1 minus.

 

[pka]

The combination of 360 choosing 10 is SEE HERE. and 400 choose 10 is SEE HERE
The ratio is SEE HERE That ratio is the probability that none of your ten match the one.
So the complement, \(1-r\) is the probability of at least one.

 

[lordanyc]

thank you. so basically exactly what I wrote in my prior post? 66% is probability that at least one out of 10 cards will match 1 random statement card? Sorry for being repetitive but want to be clear as sometimes a word can change a meaning