Q. 8.19 - A First Course in Probability

Question

A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y denote the number of fish that need be caught to obtain at least one of each type.

(a) Give an interval (a, b) such that $P(a\le Y\le b)\ge 0.9$

(b) Using the one-sided Chebyshev inequality, how many fish need we plan on catching so as to be at least 90 percent certain of obtaining at least one of each type?

Step-by-Step Solution

Answer

(a.) The interval is $(4,16)$

(b.) $0.968325$

Given information

Their are 4 different types of fishes.

Y denotes the no. of fishes that need to be caught to obtain atleast one of each type.

Let p denotes the probability of catching fish of atleast one type.

Formulation

Since this is a geometric distribution with parameter p. 

Let $A, B, C \mathrm{and} D$ denotes the event of drawing fishes of different types.

Therefore,

$p=P\left(A\cup B\cup C\cup D\right)$

Calculating p

$$\begin{gathered} \Rightarrow p=P\left(A\right)+P\left(B\right)+P\left(C\right)+P\left(D\right)-P(A\cap B)-P(A\cap C)-P\left(A\cap D\right)- \\ P(B\cap C)-P\left(B\cap D\right)-P\left(C\cap D\right)-P\left(A\cap B\cap C\right)-P\left(A\cap B\cap D\right)-P\left(A\cap C\cap D\right) \\ -P(B\cap C\cap D)+P\left(A\cap B\cap C\cap D\right) \end{gathered}$$

Since the events are independent, therefore,

$P(A\cap B)=P\left(A\right)P\left(B\right)=\frac{1}{4}*\frac{1}{4}=\frac{1}{16}$

This applies to each terms. Therefore,

$$\begin{gathered} p=4*\frac{1}{4}-6*\frac{1}{16}+4*\frac{1}{64}-\frac{1}{256} \\ \Rightarrow p=\frac{175}{256} \end{gathered}$$

Part (a)

We need to find interval $(a,b)$ such that $P(a\le Y\le b)\ge 0.9$

$P(X\le 2)\approx 0.96832$

Therefore, $1\le X\le 2$

Therefore the no. of draws will be in between $(4^1,4^2)$ ie. $(4,16)$

So, $a=4 \mathrm{and} b=16$

Part (b)

We need to find Chebychev's inequality in order to be $90\%$ certain that we get atleast one of each type.

$P\left(\right|X-𝜇|<k𝜎)\ge 1-\frac{1}{k^2}$

In case of Geometric distribution, we have

$$\begin{gathered} 𝜇=\frac{1}{p}=\frac{256}{175} \\ 𝜎=\frac{q}{p^2}=\frac{20736}{30625} \end{gathered}$$

Since RHS = 0.9. Therefore in above inequality,

$$\begin{gathered} 1-\frac{1}{k^2}=0.9 \\ \Rightarrow k=\sqrt{10} \end{gathered}$$

Calculations

So, the above inequality simplifies to 

$$\begin{gathered} P\left(\left|X-\frac{175}{256}\right|<\frac{20736\sqrt{10}}{30625}\right) \\ =P\left(\left|X-0.6836\right|<2.1412\right) \end{gathered}$$

Now instead of calculating right and left hand inequality, we just calculate the right hand inequality and that should cover atleast 45% of the area.

$$\begin{gathered} =P(X<2.8248) \\ =P(X\le 2) \end{gathered}$$

Therefore,

$P\approx 0.968325$