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I have three marbles in a bag. Red, Yellow and Blue. If I reach into the bag without looking and randomly pick one out, and then place it back into the bag, and repeat this 15 times, what are the odds that I will not get a blue one a single time? So out of all the tries what are the odds of it not happening?

2006-07-11 03:43:25 · 5 answers · asked by Mephisto 2 in Science & Mathematics Mathematics

5 answers

Here's how we calculate something like this.

What is the probability of not getting blue on the first draw? 2/3, since there are 2 non-blue marbles out of the three in the bag.

What is the probability of not getting blue on the second draw? 2/3, since there are 2 non-blue marbles out of the three in the bag.

What is the probability of not getting blue on the third draw? 2/3, since there are ... you get the idea. For any single draw, the probability of not getting a blue marble is 2/3.

Now when we have a repeated experiment, the wording (AND or OR) is important. This this case, we are working with AND. What we are looking for is the following:

(no blue on first draw) AND (no blue on second draw) AND (no blue on third draw) AND ... AND (no blue on sixteenth draw).

The word AND in this means that we multiply the probabilities. So the solution is found by

2/3 x 2/3 x 2/3 x 2/3 x ... x 2/3
= (2/3)^16 {2/3 to the sixteenth power}
= 65536/43046721
= 0.00152243884

2006-07-11 13:56:09 · answer #1 · answered by tdw 4 · 0 0

In probability, when you have two different quantities and you want to say "x AND y", you use multiplication, whereas for "x OR y", you use summation (assuming independance of the variables).

So in your case, we actually only have one quantity, repeated N=15 times, and you want to say "what's the probability of x AND x AND x ..." et cetera.

So, we know that our finaly answer will just be x^15 (x to the power of 15, or x times itself fifteen times).

Now all we need to know is this pesky 'x'. Well, for a _single_ pull from the bag, there is a 2/3 chance that we won't draw a blue (in other words, two of the three equally likely draws result in non-blue marbles).

So our final answer is: *drum roll*: (2/3)^15, which a probability of about 0.002.

2006-07-11 03:55:16 · answer #2 · answered by Michael C 1 · 0 0

The chance of not drawing a blue marble on any one try would be 2 in 3, or 0.66666....

The chance of not drawing a blue marble on 16 consecutive tries would then be (0.6666....)^16 = 0.001522, astoundingly low odds.

2006-07-11 06:05:23 · answer #3 · answered by Kyrix 6 · 0 0

(2/3)^15

2006-07-11 03:51:01 · answer #4 · answered by gjmb1960 7 · 0 0

I dunno. But i bet you never get a green one.

2006-07-11 04:16:39 · answer #5 · answered by bequalming 5 · 0 0

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