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Mathematics 13 Online
OpenStudy (anonymous):

which pair has equally likely outcomes? a standard deck of cards has 12 face cards and 4 aces. what is the probability of drawing a black 2? drawing a 6? rolling a sum of 7 on two fair six-sided dice? rolling a sum of 3 on two fair six-sided dice? rolling a sum of 10 on two fair six sided dice?

OpenStudy (anonymous):

Unfortunately, I see no 2 event with the same outcome here. (I can explain in detail how I got to each outcome if you wish) 1st one is 2/52, so 1/26 2nd one is 4/52, so 1/13 3rd one is 6/36, so 1/6 4th one is 2/36, so 1/18 and finally, last one is 3/36, so 1/12

OpenStudy (anonymous):

yes, please advise how you came to these outcomes.

OpenStudy (anonymous):

sure, ok, for the first 2, it's fairly simple. The possibility of drawing a black 2 out of a 52 card deck (I assume it's a 52 card deck) is of 2 out of 52 different cards. Only the 2 of clubs and of spades are the carsd you wish. For drawing a 6, well, there are 4 different 6s in a deck, so there are 4 chances out of 52 that you'll get the wanted result The next 3 a bit trickier (I'll keep going on the next post)

OpenStudy (anonymous):

ok thank you for your help i really appreciate it

OpenStudy (anonymous):

For the possibility of getting a sum of 7 on 2 dice rolls, you need to find out what are the different possibilities to look for : 1 and 6 2 and 5 3 and 4 4 and 3 5 and 2 6 and 1 All these have a sum of 7. So, now, what are the odds that we get 1 and 6? it would be 1/6 * 1/6, right? Because there's 1 chance out of 6 possibilities that you'll roll a one on the first dice and a 1 chance out of 6 possibilities that you'll roll a one on the second dice. The probability of getting 2 and 5 is the same thing, 1/6*1/6. Same thing for the next 4 pairs of results. In the end, you have : (1/6*1/6)*6 = 6/36 = 1/6

OpenStudy (anonymous):

ok, i'm with you so far

OpenStudy (anonymous):

for the last 2, it's basically the same approach. For a sum of 3, what are the combinations that you want? It's either 1 and 2 or 2 and 1. So, 1/6*1/6 for 1 and 2 and 1/6*1/6 for 2 and 1, for a grand total of : (1/6*1/6)*2

OpenStudy (anonymous):

how about that last one, think you can find out why it is 3/36?

OpenStudy (anonymous):

you mean 1/12?

OpenStudy (anonymous):

so for the last one you have 2 possible combination - 5 & 5, or 4 & 6.

OpenStudy (anonymous):

so that would be 1/6 * 1/6 (2)?

OpenStudy (anonymous):

well, I see what you are doing, and it makes more sense intuitively to do this, however, probabilities are a weird thing ;-) (4 and 6) and (6 and 4) are actually not the same thing. It's actually quite hard to explain, because myself have a hard time understanding it, but these are 2 "distinct" outcomes

OpenStudy (anonymous):

SO IT WOULD BE 1/3 PROBABLITY THEN?

OpenStudy (anonymous):

not quite. Just going back to what you had written, 1/6 * 1/6 (2), the 2 in there would in fact be a 3 (because of the 3 different combinations that would satisfy the wanted condition, that is to get a sum of 10)

OpenStudy (anonymous):

SO AGAIN 1/6 * 1/6 (3) = 3/36 = 1/12

OpenStudy (anonymous):

BECAUSE IT HAS 3 POSSIBLE OUTCOMES.

OpenStudy (anonymous):

yep, exactly (your capslock is on by the way)

OpenStudy (anonymous):

THE PROBABILITY OF GETTING 4 AND 6 IS 1/3, OF GETTING 6 AND 4 IS 1/3 AND 5 AND 5 IS 1/3 to summarize further.

OpenStudy (anonymous):

oh, I see what you mean! Yes, they all account for 1/3 of the wanted outcome

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

or is this going to far with it?

OpenStudy (anonymous):

well, yes and no. I mean, it's as good as any way of seeing it, if you understand it better that way, it's fine. It,s just that, getting to the point where you find that the probability that (4 and 6) are 1/3 of what you are looking for, you ahave to find the "3" first and you could just stop there. for the context of this specific problem, it may be taking it a bit too far, but for other problems, it's a better way of seeing it :-) It's what you understand it best with that counts in the end

OpenStudy (anonymous):

hello? is that going too far with it?

OpenStudy (anonymous):

oh, ok thank you for your help!

OpenStudy (anonymous):

not a problem!

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