doturitip9

2022-07-14

There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.

I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12}\cdot \frac{6}{11}\cdot \frac{5}{10}=\frac{210}{1320}$
b) $1-\left(\frac{7}{12}\cdot \frac{6}{11}\cdot \frac{5}{10}+\frac{5}{12}\cdot \frac{4}{11}\cdot \frac{3}{10}\right)=\frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?

kawiarkahh

Expert

(a)
$\frac{\left(\genfrac{}{}{0}{}{7}{3}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}$
You can show that this equals what you did.
$\begin{array}{rl}\frac{\left(\genfrac{}{}{0}{}{7}{3}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}& =\frac{\frac{7!}{3!4!}}{\frac{12!}{3!9!}}\\ \\ & =\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}\frac{3\cdot 2\cdot 1}{12\cdot 11\cdot 10}\\ \\ & =\frac{7}{12}\cdot \frac{6}{11}\cdot \frac{5}{10}\end{array}$
(b)
$1-\frac{\left(\genfrac{}{}{0}{}{7}{3}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}-\frac{\left(\genfrac{}{}{0}{}{5}{3}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}$

Wronsonia8g

Expert

a)
$\frac{\left(\genfrac{}{}{0}{}{7}{3}\right)\left(\genfrac{}{}{0}{}{5}{0}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}$
b)
$1-\frac{\left(\genfrac{}{}{0}{}{7}{3}\right)\left(\genfrac{}{}{0}{}{5}{0}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}-\frac{\left(\genfrac{}{}{0}{}{7}{0}\right)\left(\genfrac{}{}{0}{}{5}{3}\right)}{\left(\genfrac{}{}{0}{}{12}{3}\right)}$

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