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?

Short answer

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

(b.)$0.968325$

Step-by-step solution

Given information

Their are 4 different types of fishes.

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

Let pdenotes 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.

role="math" localid="1650538846114" $P\left(\right|X-𝜇|<k𝜎)\ge 1-\frac{1}{k^2}$

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,

$P\approx 0.968325$