Jellybean question

intrakate1

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Feb 1, 2011
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If i have a jar with 15 red, 13 yellow, and 7 green jelly beans what is the probability that i will pull out 1 red and 2 yellows (without the replacement of the ones i removed)
 
Hello, intrakate1!

If i have a jar with 15 red, 13 yellow, and 7 green jelly beans,
what is the probability that i will pull out 1 red and 2 yellows
(without the replacement of the ones i removed)?

If you mean Red- ⁣Yellow- ⁣Yellow in  that  order, the probability is:   153513341233  =  781309\displaystyle \text{If you mean Red-\!Yellow-\!Yellow }in\;that\;order\text{, the probability is: } \;\frac{15}{35}\cdot\frac{13}{34}\cdot\frac{12}{33} \;=\;\frac{78}{1309}


If you mean 1 red, 2 yellows in  any  order, the probability is:   (151)(132)(353)  =  11706545  =  2341309\displaystyle \text{If you mean 1 red, 2 yellows }in\;any\;order\text{, the probability is: }\; \dfrac{{15\choose1}{13\choose2}}{{35\choose3}} \;=\;\frac{1170}{6545} \;=\;\frac{234}{1309}

 
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