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anonymous
 one year ago
A bag contains eight American League baseballs and five National League baseballs. Umpire Strichsbach reaches in and selects three balls without looking.
A. What is the probability that all three are American League Baseballs?
B. What is the porbability that all three are National League Baseballs
anonymous
 one year ago
A bag contains eight American League baseballs and five National League baseballs. Umpire Strichsbach reaches in and selects three balls without looking. A. What is the probability that all three are American League Baseballs? B. What is the porbability that all three are National League Baseballs

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you want to add the cards up so you know how many you have in total

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. 13 balls combined at initially, 8 of them are AL. So the probability of selecting an AL ball with the first selection is 8/13. After the first selection, 12 balls are left, 7 of them are AL. The prob. of selecting and AL ball is 7/12 After the 2nd selection, 11 balls are left. 6 of them are AL. The prob. of selecting an AL ball is 6/11. Multiply them to get the combined probability \[\frac{ 8 }{ 13 }\times \frac{ 7 }{ 12 }\times \frac{ 6 }{ 11 }\] Do the same for NL balls

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0A less wellknown solution is to use the hypergeometric distribution, which is \(\large \frac{\left(\begin{matrix}A \\ a\end{matrix}\right)\left(\begin{matrix}N \\ n\end{matrix}\right)}{\left(\begin{matrix}A+N \\ a+n\end{matrix}\right)}= \frac{\left(\begin{matrix}8 \\ 3\end{matrix}\right)\left(\begin{matrix}5 \\ 0\end{matrix}\right)}{\left(\begin{matrix}8 \\ 3\end{matrix}\right)}=\frac{28}{143}\) A=initial number of American League balls a=number of American League balls drawn N=initial number of National League balls n=number of National League balls drawn To solve the second part, put a=0, n=3
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