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# A simple probability problem...

Da Rossa
Which I never learnt how to solve in high school back then, I'm ashamed of it...

Hi, what about this: a deck of 40 cards contains 23 blue and 17 red cards. I randomly pick up 7 cards. What are the odds:
1 - I take zero red cards;
2 - I take exactly three red cards;
3 - I take exactly seven cards?

Thanks
Bikerman
....
kelseymh
 Da Rossa wrote: Which I never learnt how to solve in high school back then, I'm ashamed of it... Hi, what about this: a deck of 40 cards contains 23 blue and 17 red cards. I randomly pick up 7 cards. What are the odds: 1 - I take zero red cards; 2 - I take exactly three red cards; 3 - I take exactly seven cards? Thanks

The easiest way to show you how to solve these is brute force. There are numerous formulas you could memorize (look up choose notation in your favorite math resource), but if you know how to do it, you can derive the notation and formulas for any future problem.

The basic thing to know is that probabilities multiply. If the chance of A happening is 50%, and the chance of B happening is 30%, then the chance of both A and B happening is 0.5*0.3 = 0.15, or 15%.

So, for part 1, step through drawing each card individually. The probability for drawing zero red cards is the same as for drawing all blue. For the first card, you've got 23 blue out of 40, so P(B1) = 23/40. For the second card, there are 22 blue out of 39 cards left, so P(B2) = 22/39. For the third card, P(B3) = 21/38, and so on. The total for zero red cards in a draw of seven will be P = (23*22*21*20*19*18*17)/(40*39*38*37*36*35*34) = 0.013.

I'll leave the other two for you to work out.
Da Rossa
Sorry, I did take a brief look at your reponses but had to leave to work that day. So I got no more notifications for this topic,
I'll look better soon. Thanks Biker and kelseymh!
Arrogant
These are some simple problems of combination an permutations
Just use combination and probability

probability = no of favorable cases/ total no of cases
Combination C(n,r) = n!/((n-r)!*r!)

For the first question, you get zero red cards,so all of em are blue
1. C(23,7)/C(40,7)
For second ,you get 4 blues and 3 reds
2. C(23,4)*C(17*3)/C(40,7)
For third
3. You assumed that you took 7 cards so the probability would be one ofcourse

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