probablity problem

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i had an argument with my teacher today if anyone can help id appreciate it

A bag contains 6 red balls, 4 white balls and 5 blue balls. Two balls are drawn from the bag. Find the probability that none is blue

my answer is up the top in black next to ii (iii, iv and v are for different questions) the instructor's fi1's anser is in red on the right and his seccond is in blue at the bottom

http://i174.photobucket.com/albums/w94/ ... an0001.jpg

i took all the possible solutions (1) and took away the probability that the first one would be blue, the probability of white then blue and red then blue, the solution is .4217687075


this stuff isn't essential

the instructor apparently took the probability the both would be blue and for some reason multiplied to by the probability of getting one of each. it gives .0453514739, if you add them if gives .5714285714

his second is apparently the ways neither could be blue over all the possible solutions, gives .4285714286

he kept on saying i wasn't taking into account both balls being blue but the probability that the first one is blue includes that, i drew a tree diagram to try and explain but i dident get to use it on him

http://i174.photobucket.com/albums/w94/ ... an0002.jpg

Hi. I did the problem without reading either of your answers, as not to be biased. Here is how I understood the problem. This is important because many of these disputes arise because of a misunderstanding of the original problem.

A bag has 5 blue balls, 6 red balls, and 4 white balls. You reach in, pull out a ball, and take it out. Then you reach in, pull out another ball, and take it out. What is the chance that none of the balls you pulled are blue?

The answer is: Chance of pulling a non-blue ball on the first pull * Chance of pulling a non-blue ball on the second pull given that the first event is true.

Or: 10/15 * 9/14 = 42.86%.

So, your professor is right. I will try to read your answer and see what's wrong with it now.

My credentials: I have a BS in CS and I was a professional gambler for 2 years.

You computed the probability as 1 - 5/15 - (6/15)(5/14) - (4/15)(5/14) = 1 - 1/3 - 1/7 - 2/21 = 21/21 - 7/21 - 3/21 - 2/21 = 9/21 = 3/7 != .4217687075

3/7 is the answer; your methods are correct (though your conversion to decimals isn't).

Every approach to this problem yields the same result - treating it as two trials we get (2/3)(9/14) = 3/7; (10C2)/(15C2) = 3/7. I don't see what your teacher's issue is. I don't even see what he's doing.

(PS: probability and combinatorics are my worst subjects...)

Oh ya, twoshots is right. Your answer is fine, you just didn't convert to decimal properly.

The way I would do it is sequentially asking what the probability is at each stage of drawing a non-blue ball.

There are 10 non blue and 5 blue, for 15 balls total. The probability of drawing a single non-blue is 10/15 = 2/3

Once a non-blue ball is drawn, there are 9 non-blue balls and 5 blue balls left for 14 balls total. The probability of drawing a single non-blue ball is then 9/14.

Then the total probability of both events occuring is (2/3)*(9/14) = 0.42857142

edit: yikes, you guys are fast

thanks a lot that was confusing the hell out of me

for some reason im having trouble getting my head around C and P even though ive done it twice before, also i cant understand what that teacher is talking about half the time, wich might become a big problem cus i have him for 2 classes that cost about $1500 a semester

C represents how many different ways you can choose a couple of elements out of a set.

So there's 15 different balls and you choose two.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

If you were to write out all of the different ways you could choose two balls, it'd be like: 1 and 2, 1 and 3, 1 and 4, 1 and 5.. really long list here ... 14 and 15.

How do you found out how long that list is? The number of elements in that list is 15 C 2. The C represents how many ways there are to "choose" 2 elements out of the list of 15.

So let's say that 11-15 are the blue balls, and 1-10 are the non-blue balls. How many different ways can you pick two non blue balls?

1 2 3 4 5 6 7 8 9 10

The list would be.. 1 and 2, 1 and 3, 1 and 4, 1 and 5 ... long list here ... 9 and 10. This would be just like the other list, except any pair containing a blue ball isn't valid here. 4 and 13 is in the other list, but not this one. So this one is shorter.

So how long is this list? 10 C 2.

What's the general formula for probabilty of an outcome? It's valid outcomes / possible outcomes.

So here, you can use 10 C 2 / 15 C 2. Which is what your teacher did.