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magley64

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4/13. Based on what i just read.

So a simple probability problem that is very common would be to ask what is the probability of pulling out a 6. This probability would be 4/52 which could then be simplified to 1/13. Notice there are 4 six’s in a deck out of a total of 52 cards which is how we get 4/52. You should always simplify your final answer when expressing a probability.

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4/13. Based on what i just read.

So a simple probability problem that is very common would be to ask what is the probability of pulling out a 6. This probability would be 4/52 which could then be simplified to 1/13. Notice there are 4 six’s in a deck out of a total of 52 cards which is how we get 4/52. You should always simplify your final answer when expressing a probability.

That's how I came up with 1 in 3.25. What does the Prof say?

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he's defending 17/52 because there are 4 aces "plus" 13 diamonds....

doesn't sound to me like he should be teaching anyone statistics...:rolleyes:

he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only?

i mean if what is the probability we will pick a diamond = 13 .

what is the probability we will pick an ace = 4

then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.?

then again maths is not my strong subject..:confused:

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he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only?

i mean if what is the probability we will pick a diamond = 13 .

what is the probability we will pick an ace = 4

then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.?

then again maths is not my strong subject..:confused:

52 cards in a deck, only 16 cards in that deck are a successful pick, odds are 16/52 broken down to 4/13. Replace it or not there are still only 16 cards out of 52 that are successful picks. Not to mention not putting the card back after a draw would change the odds with every draw.

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i mean what if the argument is like this, first time i picked a diamond ACE i am going to count it as diamond, then i replace the card and then next time i pick it again and i count it as ACE. until the card reaches our hand its not counted, so in the deck of 52 cards this one card hold value of two probability ?

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he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only?

i mean if what is the probability we will pick a diamond = 13 .

what is the probability we will pick an ace = 4

then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.?

then again maths is not my strong subject..:confused:

I think this is the professor's line of thinking...

Try this: put an X on 16 cards, and then ask "what are the odds of drawing a card with an X?"

your answer would be 16/52, right?

now draw an X on the 13 diamonds, and the 3 remaining aces, the cards with the X, are the same cards referenced in the original problem, but now there is only 1 classification (whether or not the card has "X" drawn on it)

I gotta believe 16/52 is the only correct answer

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now ur suggesting the probability of picking a card marked X which becomes 16/52

now lets do this, put an x on all diamonds, what is the probability of picking card marked x = 13/52

now put a Y on all aces probability is 4/52

what is the probability that of picking either x or y ... i pick back to back 16 x put it back then back to back 4 y then it is 17, the condition never excluded a thing with both x and y.

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