Question 326460: I would like to see the mechanics of how to figure out the formula for this so I can do this myself. I have been checking out Permutations at various websites, my efforts have not been fruitful, but I have tried.
You have a deck of 40 cards of different colors. The deck contains 3 red cards and 3 blue cards. The other 34 cards are not red or blue. The deck is randomly shuffled.
1) What is the probability of drawing at least 1 red and 1 blue card in the first 6 cards drawn from the top of the deck?
2) What is the probability of drawing at least 1 red and 1 blue card in the first 10 cards drawn from the top of the deck?
Thank you. Ken
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
This is one of those cases where it will be easier to calculate the probability of utter failure to get what you want and subtract that from 1. We will need the probability that none of the cards are red plus the probability that none of the cards are blue minus the probability that none of the cards are red or blue.
Let A = the event that at least one card is Red. Let B = the event that at least one card is Blue. Then A or B is the event that at least one card is either red or blue. ~A means no card is red, ~B means no card is blue, and ~(A or B) means ALL of the cards are something other than red or blue.
We need to calculate \ +\ P(\neg{B})\ -\ P\left(\neg(A\,or\,B)\right)\right))
We need to subtract P(~(A or B)) to avoid double counting since P(~A) and P(~B) are NOT mutually exclusive events.
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and is calculated by !})
First we want the probability that NONE of the first 6 cards are RED: 6 trials, 0 successes, probability of success  for No REDs:
\ =\ \left(6\cr 0\right\)\left(\frac{3}{20}\right)^0\left(\frac{17}{20}\right)^6)
The arithmetic is simpler than it looks because \ =\ 1) always, and anything raised to the zero power is 1, so you really only have to compute ^6)
Next we want the probability that NONE of the first 6 cards are BLUE: Again, 6 trials, 0 successes, probability of success for No BLUEs:
It is the same as the number for no reds.
Last we want the probability that NONE of the first 6 are RED or BLUE: 6 trials, 0 successes, probability of success  for No REDs or BLUEs:
\ =\ \left(6\cr 0\right\)\left(\frac{6}{20}\right)^0\left(\frac{14}{20}\right)^6)
But all you need to calculate is ^6)
Putting it all together:
We need to calculate
which, in terms of something that will make sense to our calculator is: ^6\ +\ \left(\frac{17}{20}\right)^6\ -\ \left(\frac{14}{20}\right)^6\right))
Just break out the calculator and start punching, or, if you have Excel or something like it handy, open a spreadsheet and set it up to do the calculations for you.
John
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