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How Would I Solve this math Problem?
Just to put this out there, I'm a High School student, i've taken up to PreCalc level math, and I'm taking Calc next year. My question is, how would I set up a formula to answer something like this:
Lets say I had a normal deck of cards (52 without jokers), and I wanted to solve the probability of drawing a 3 on the top. Now clearly that's a fraction of 4/52 (or 1/13). But what is the probability that I would draw a 3 within the first 7 draws (and once a card has been drawn, the total number of the deck is reduced by 1)?
Do I just add together each individual fraction and take the average, or what? If there is a simple formula for it, let me know.
4 Answers
- 1 decade agoFavourite answer
Actually it is much simpler to turn the problem around and compute the probabitlity of not drawing any 3's in the first seven cards. Then subtract that prob from 1.
P(no three in first draw) = 48/52
P(no three in 2nd draw) = 47/51
o
o
o
P(no three in 7th draw) = 42 / 46
P(no three in 7 draws) = (48x47x46x45x44x43x42) divided by (52x51x50x49x48x47x46)
P(no three in 7 draws) =
371090522880 / 674274182400 = 0.550355527
finally p(at least one three in seven draws) =
1 - 0.550355527 = 0.449644473
Have fun in Calc, but instead of looking for simple formulas, you will be better off trying to understand the logic behind this kind of approach to probability problems.
- Andy JLv 71 decade ago
Probability is a tricky subject, and there is generally more than one correct way to approach the problem. It is also tricky in that if you do not define the problem fully, you can get a different answer depending on how you approach the problem.
In this case, you have not fully defined the problem. Are you asking what's the probability of drawing at least one 3 within the first 7 cards? Or exactly one 3 within the first 7 cards? Or do you stop as soon as you draw a 3? Each of these three problems has a different answer.
The first one:
You are drawing 7 cards, and want at least one of them to be a 3. There are 52C7 = 133,784,560 ways to draw 7 cards. The ways to draw no 3's are 48C7 = 73,629,072. So the ways to draw at least one 3 is 52C7 - 48C7 = 60,155,488. The probability of this happening is 60155488/52C7 = 3478/7735.
The second one:
You are drawing 7 cards, and want exactly one of them to be a 3. There are 52C7 = 133,784,560 ways to draw 7 cards. The ways to draw exactly one 3 is 4 * 48C6 = 49,086,048. That's because there are 4 ways to choose a 3, and 48C6 ways to choose 6 from 48 non-threes. The probability is 2838/7735.
The third one is much trickier as you have to add up 7 cases where the 3 is the first card, the second, and so on. You have the answer for the first card, 1/13. For the second card, since we know you have not drawn the 3 on the first card, there are still 4 threes, but only 51 cards left. So the probability of getting a non-3 followed by a 3 is 48/52 * 4/51 = 16/221. Similarly, the probability of two non-3s followed by a 3 is 48/52 * 47/51 * 4/50. Keep going up to 7 and add them up for the final probability.
- TychaBraheLv 71 decade ago
The possibility of not drawing a 3 on the first card is 48/52.
The possibility of not drawing a 3 on the second card is 47/51. And so on.
(48 * 47 * 46 * 45 * 44 * 43 * 42) / (52 * 51 * 50 * 49 * 48 * 47 * 46)
The answer is just over 55%.
So the probability of drawing a 3 would be just over 45%.
