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# Math

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A guy own 10 white toy bulls, 14 yellow toy bulls, and 8 black toy bulls.

If he is blindfolded, and puts 11 bulls in one bag, and 11 in another bag, what is the chance of a yellow bull being in bag 1, and the chance of a black bull being in both bag 1 and 2?

They're are 10 leftover bulls. What is the chance of a white bull being in the group of leftover bulls?

Nov 27, 2018
edited by DaP1ckle  Nov 27, 2018

#1
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$$P[\text{at least 1 yellow bull in bag 1}]=\\ 1-P[\text{no yellow bulls in bag 1}] = \\ 1 - \dfrac{\sum \limits_{w=3}^{10}\dbinom{10}{w}\dbinom{8}{11-w}}{\dbinom{32}{11}}=\dfrac{206719}{206770}$$

I'll post #2 when I figure it out

$$P[\text{white bull in group of leftovers}] = \\ 1 - P[\text{all white bulls in bag 1 or bag 2}] = \\ \text{(think of bag 1 and bag 2 as one big bag here)} \\ 1 - \dfrac{\sum \limits_{y=4}^{12}~\dbinom{14}{y}\dbinom{8}{12-y}}{\dbinom{32}{22}}=\dfrac{2456369}{2481240}$$

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Nov 30, 2018
#2
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P[\text{at least 1 black bull in each bag}] = \\ 1 - P[\text{no black bull in bag 1 or no black bull in bag 2 or both}] = \\ \begin{align*} 1 \\ &- P[\text{no black bull in bag 1}]\\ &-P[\text{no black bull in bag 2}]\\ &+P[\text{no black bull in either bag}] \end{align*}

$$\text{As the problem is symmetric w/respect to the bags}\\ P[\text{no black bull in bag 1}] = P[\text{no black bull in bag 2}]\\ P[\text{no black bull in bag 1}] =\\ \dfrac{\sum \limits_{w=0}^{10}~\dbinom{10}{w}\dbinom{14}{11-w}}{\dbinom{32}{11}} =\dfrac{2261}{116870}$$

$$P[\text{no black bull in either bag}] = \\ \dfrac{\sum \limits_{w=0}^2~\dbinom{10}{w}\dbinom{14}{2-w}}{\dbinom{32}{10}}=\dfrac{1}{233740}$$

$$P[\text{at least 1 black bull in both bags}] = \\ 1-2\dfrac{2261}{116870}+\dfrac{1}{233740} = \dfrac{224697}{233740}$$

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Dec 1, 2018