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Question Number 111163 by Aina Samuel Temidayo last updated on 02/Sep/20

$$\mathrm{A}\mathrm{bag}\mathrm{contains}11\mathrm{white}\mathrm{balls}\mathrm{and}9$$$$\mathrm{black}\mathrm{balls},\mathrm{another}\mathrm{contains}12\mathrm{white}$$$$\mathrm{balls}\mathrm{and}13\mathrm{black}\mathrm{balls}.\mathrm{If}\mathrm{two}\mathrm{balls}$$$$\mathrm{are}\mathrm{drawn}\mathrm{without}\mathrm{replacement},\mathrm{from}$$$$\mathrm{each}\mathrm{bag}\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}$$$$\mathrm{exactly}\mathrm{one}\mathrm{of}\mathrm{the}4\mathrm{balls}\mathrm{is}\mathrm{white}?$$

Commented by Aina Samuel Temidayo last updated on 02/Sep/20

$$\frac{\left(\begin{array}{c}11\\ 1\end{array}\right)\times \left(\begin{array}{c}9\\ 1\end{array}\right)}{\left(\begin{array}{c}20\\ 2\end{array}\right)}\times \frac{\left(\begin{array}{c}12\\ 0\end{array}\right)\times \left(\begin{array}{c}13\\ 2\end{array}\right)}{\left(\begin{array}{c}25\\ 2\end{array}\right)}+\frac{\left(\begin{array}{c}11\\ 0\end{array}\right)\times \left(\begin{array}{c}9\\ 2\end{array}\right)}{\left(\begin{array}{c}20\\ 2\end{array}\right)}\times \frac{\left(\begin{array}{c}12\\ 1\end{array}\right)\times \left(\begin{array}{c}13\\ 1\end{array}\right)}{\left(\begin{array}{c}25\\ 2\end{array}\right)}$$$$=\frac{117}{500}$$

Commented by Aina Samuel Temidayo last updated on 02/Sep/20

$$\mathrm{Is}\mathrm{this}\mathrm{a}\mathrm{correct}\mathrm{solution}\mathrm{to}\mathrm{the}$$$$\mathrm{problem}?$$

Answered by john santu last updated on 02/Sep/20

Commented by Aina Samuel Temidayo last updated on 02/Sep/20

$$\mathrm{Yes}.\mathrm{Thanks}.$$

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