Q.1.34 - A First Course in Probability

Question

Suppose that $10$fish are caught at a lake that contains $5$distinct types of fish.

$\left(a\right)$How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the $5$types?

$\left(b\right)$ How many outcomes are possible when$3$ the $10$fish caught are trout?

$\left(c\right)$ How many when at least $2$of the $10$are trout?

Step-by-Step Solution

Answer

The answers are $1001,120$and $495$respectively.

Step 1 Given Information.

Given that$10$ fish are caught at a lake that contains $5$distinct types of fish.

Step 2 Explanation.

There are $5$types of fish in the lake and precisely $10$fish are caught at a lake. We are interested in a number of possible outcomes, where we'll define outcome as a number of caught fish of each of the $5$types.

Let's$t_1,t_2,t_3,t_4,t_5$ define the number of each of the 5 types of fish caught at the lake.

Obviously, we have: $t_1+t_2+t_3+t_4+t_5=10$

Now, this becomes a standard problem of finding the number of solutions with constraints$t_k\in \{0,1,\dots ,10\}$. We can solve it in the following way: consider $14$undistinguishable balls placed consecutively. From the given$14$ balls we will choose $4$and these $4$balls will define a unique partition.

The number of partitions is the same as the number of solutions to the equation. Thus we have precisely:

$\left(\begin{array}{c}14\\ 4\end{array}\right)=\frac{14·13·12·11}{1·2·3·4}=1001$

Hence, there are$1001$ different possible outcomes.

Now if we consider that $3$out of the$10$ caught fish are trout then there are:

$\left(\begin{array}{c}10\\ 3\end{array}\right)=\frac{10·9·8}{1·2·3}=120$

possible outcomes.

If there are at least $2$trouts in the $10$caught fish to calculate the number of possible outcomes we will use the inclusion-exclusion principle. We know that the number of possible outcomes is$1001$. To find out what is the number of outcomes with at least$2$ trout we'll just subtract outcomes where there is$1$ trout or no trout at all.

There are preciseness$\left(\begin{array}{c}13\\ 3\end{array}\right)=286$ outcomes with no trout and$\left(\begin{array}{c}12\\ 3\end{array}\right)=220$ outcomes with 1 trout caught.

Therefore there must be$1001-286-220=495$outcomes where at least$2$ trout are caught. Hence, we are done.

Step 3 Conclusion.

The answers are$1001,120$ and $495$respectively.