Chapter 13.5, Problem 30E

To determine

To calculate: The value of probability $P\left(\text{not walleye, given trout and perch}\right)$ when three fish are removed from an aquarium.

Answer to Problem 30E

The value of probability $P\left(\text{not walleye, given trout and perch}\right)$ when three fish are removed from an aquarium is $\frac{3}{4}$ .

Explanation of Solution

Given information:

There are 6 fish in an aquarium namely a trout, a bass, a perch, a catfish, a walleye and a salmon and three fish at random are drawn.

Formula used:

Probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information thatthere are 6 fish in an aquarium namely a trout, a bass, a perch, a catfish, a walleye and a salmon and three fish at random are drawn.

The probability of $P\left(\text{not walleye, given trout and perch}\right)$ .

Let A be an event that removed fish is not walleye and B be an event that removed fish is trout and perch.

Use combination to evaluate the probability, recall that if there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Recall that probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that removed fish is trout and perch,

  $\begin{array}{c}P\left(B\right)=\frac{C\left(1,1\right)\cdot C\left(1,1\right)\cdot C\left(4,1\right)}{C\left(6,3\right)}\\ =\frac{\frac{4!}{1!3!}}{\frac{6!}{3!3!}}\\ =\frac{4}{20}\end{array}$

Probability that removed fish is not walleye but can be trout and perch,

  $\begin{array}{c}P\left(A\text{ and B}\right)=\frac{C\left(1,1\right)\cdot C\left(1,1\right)\cdot C\left(3,1\right)}{C\left(6,3\right)}\\ =\frac{\frac{3!}{1!2!}}{\frac{6!}{3!3!}}\\ =\frac{3}{20}\end{array}$

Recall that given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability of $P\left(\text{not walleye, given trout and perch}\right)$ .

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{3}{20}}{\frac{4}{20}}\\ =\frac{3}{4}\end{array}$

Thus, the value of probability $P\left(\text{not walleye, given trout and perch}\right)$ when three fish are removed from an aquarium is $\frac{3}{4}$ .