Question 1158616
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Pulling two at the same time is exactly the same as pulling one and then pulling a second one without replacing the first one.


The probability of pulling a red ball where there were 8 red and 10 blue to start with is 8 divided by 18, which reduces to *[tex \Large \frac{4}{9}\ =\ 0.\overline{4}]


The probability of a red ball on the second draw, presuming a red one was drawn on the first draw means that you are now drawing from a collection that contains 7 red balls and 10 blue ones, so the probability is *[tex \Large \frac{7}{17}]


Because we accounted for the change in the number of successes and number of possibilities when we calculated the probability for the second draw, these two events are now independent and the overall probability is the product of the two individual probabilities.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{2R}\ =\ \(\frac{4}{9}\)\(\frac{7}{17}\)]


You can work out the arithmetic for yourself.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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