Math problem on probability
Started by TermiNadder2012
1/12?
i believe it is:
10C3 * (1/2)^3 * (1/2)^7 = .117
someone correct me if i'm wrontg
shouldn't it be something like 10!/3!7! x (1/2) to the 10th power.
(by the way (1/2)to the 3rd power x (1/2) 7th power = (1/2) to the 10th power.)
(by the way (1/2)to the 3rd power x (1/2) 7th power = (1/2) to the 10th power.)
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wth... please tell me you're asking this question out of pure curiosity and that it WON't be on the DAT!!!
ok but what if u find the probability of taking 3 pennies out of 5 which is 3/5...
then you find the probability of taking all 5 pennies out of the coins which is 5/10
then you multiply the 2 together?
you get 3/10. is that it? that was a wild guess... i really don't know how to do this...
ok but what if u find the probability of taking 3 pennies out of 5 which is 3/5...
then you find the probability of taking all 5 pennies out of the coins which is 5/10
then you multiply the 2 together?
you get 3/10. is that it? that was a wild guess... i really don't know how to do this...
I don't think this is actually that hard; this kind of problem seems to show up all the time. If you take one coin, you have a 5/10 probability of it being a penny. The next coin taken, you have a 4/9 probability of it being a penny. The final coin has a 3/8 probability of being a penny... so 5/10 * 4/9 * 3/8 = 1/12.
that is exactly what i thought...that is why i said 1/12 on my first post. i think that is the answer
multiplying by the diminishing fractions is correct except that it is different in this problem because the three coins are taken all at once not one at a time.
Ok so I think the answer is 10C3(p)^R(1-P)^(N-R)
R=number of success
N= Total coins
I have P=1/2 because there are 5 pennies and 5 dimes so there is a 50% chance of getting either a penny or a dime
Since this is a combination problem I have R! (which is three, the number of successes) and (N-R)!
Which would become 10!/3!7! * (1/2)^3* (1/2)^7 which gets you 3/256
R=number of success
N= Total coins
I have P=1/2 because there are 5 pennies and 5 dimes so there is a 50% chance of getting either a penny or a dime
Since this is a combination problem I have R! (which is three, the number of successes) and (N-R)!
Which would become 10!/3!7! * (1/2)^3* (1/2)^7 which gets you 3/256
I'm pretty sure 1/12 is right. Here's what I did:
Find the possible combinations of drawing 3 pennies out of the 5 pennies.
nCr: n!/[(n-r)!(r!)] 5C3: 5!/[(5-3)!(3!)] = 5x4/2x1 = 10. So there are 10 possible combinations of 3 pennies.
Now find the possible compinations of drawing any 3 coins from the 10 coins.
10C3 = 10!/[(10-3)!(3!)] = 10x9x8/3x2 = 120
possible combinations of 3 pennies
possible combinations of any 3 sets of coins
10/120 = 1/12
This is the same answer as doing 5/10 x 4/9 x 3/8 I think you can do both methods for these problems, but I'm not sure.
Remember: use the formula for combinations when the order (or arrangement) does not matter. If the order does matter, use: n!/(n-r)!
Find the possible combinations of drawing 3 pennies out of the 5 pennies.
nCr: n!/[(n-r)!(r!)] 5C3: 5!/[(5-3)!(3!)] = 5x4/2x1 = 10. So there are 10 possible combinations of 3 pennies.
Now find the possible compinations of drawing any 3 coins from the 10 coins.
10C3 = 10!/[(10-3)!(3!)] = 10x9x8/3x2 = 120
possible combinations of 3 pennies
possible combinations of any 3 sets of coins
10/120 = 1/12
This is the same answer as doing 5/10 x 4/9 x 3/8 I think you can do both methods for these problems, but I'm not sure.
Remember: use the formula for combinations when the order (or arrangement) does not matter. If the order does matter, use: n!/(n-r)!
navajo is correct, that's how I'd do it.
You can also do it using 5/10 * 4/9 * 3/8. It doesn't matter that you take them all out 'at once'.
You can also do it using 5/10 * 4/9 * 3/8. It doesn't matter that you take them all out 'at once'.
It's not different... you are still limited to taking out physical coins. If it were another problem, it could potentially matter that they're "all at once",but for the purposes of this problem, it's exactly the same.
thanks you guys for all the input; probability is something I'm not too familar with.
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