Q.7.29 - A First Course in Probability

Question

There are $4$ different types of coupons, the first $2$ of which comprise one group and the second $2$ another group. Each new coupon obtained is type i with probability $pi$, where $p1 = p2 = 1/8, p3 = p4 = 3/8$. Find the expected number of coupons that one must obtain to have at least one of

(a) all $4$ types;

(b) all the types of the first group; 

(c) all the types of the second group; 

(d) all the types of either group 

Step-by-Step Solution

Answer

a)The expected number of coupons that one must obtain to have at least one of all $4$ types is

$E\left(X\right)=\frac{437}{35}$

b)The expected number of coupons that one must obtain to have at least one of all the types of the first group is

$E\left(Y\right)=\frac{1}{2/8}+\frac{1}{1/8}=4+8=12$

c)The expected number of coupons that one must obtain to have at least one of all the types of the second group is,

$E\left(Z\right)=\frac{1}{6/8}+\frac{1}{3/8}=4$

d)The expected number of coupons that one must obtain to have at least one of all the types of either group is,

$E\left(X\right)=\frac{123}{35}$

Step 1:Given Information(part a)

Given that $4$ different types of coupons and new coupon obtained is type $i$ with probability$pi$ where $p1 = p2 = 1/8, p3 = p4 = 3/8.$

Step 2:Explanation(part a)

We are utilizing the formula from the Example $2 \mathrm{s}$. We have that

$E\left(X\right)={\int }_0^{\mathrm{\infty }} \left(1-\prod _{i=1}^4 \left(1-e^{-p_{i}x}\right)\right)dx$

we have that

$\prod _{i=1}^4 \left(1-e^{-p_{i}x}\right)={\left(1-e^{-x^2/8}\right)}^2{\left(1-e^{-3x^2/8}\right)}^2$

$=e^{-x^2}-2e^{-\left(7x^2\right)/8}+e^{-\left(3x^2\right)/4}-2e^{-\left(5x^2\right)/8}+4e^{-x^2/2}-2e^{-\left(3x^2\right)/8}+e^{-x^2/4}-2e^{-x^2/8}+1$

Integrate that over the positive real numbers to get that

$E\left(X\right)=\frac{437}{35}$

Step 3:Final Answer(part a)

$E\left(X\right)=\frac{437}{35}$

Step 4:Given Information(part b)

Given that 4 different types of coupons, the first 2 of which comprise one group and the second 2 another group and all the types of the first group.

Step 5:Explanation(part b)

Characterizing random variable $Y$ that imprints required strides to acquire numerous types from the main group. See that these means can be separated into two sections: until we have arrived at a few kinds of group one and the time until we have arrived at the excess sort. Subsequently

$E\left(Y\right)=\frac{1}{2/8}+\frac{1}{1/8}=4+8=12$

Step 6:Final Answer(part b)

$E\left(Y\right)=\frac{1}{2/8}+\frac{1}{1/8}=4+8=12$

Step 7:Given Information(part c)

 Given that all the types of the second group.

Step 8:Explanation(part c)

Also, as in part(b), the expected number of steps to get various kinds in group two is

$E\left(Z\right)=\frac{1}{6/8}+\frac{1}{3/8}=4$

Step 9:Final Answer(part c)

$E\left(Z\right)=\frac{1}{6/8}+\frac{1}{3/8}=4$

Step 10:Given Information(part d)

Given that all the types of either group. 

Step 11:Explanation(part d)

Utilize the comparative technique to some degree( a) to get that the average number of steps is equivalent to,

$E\left(X\right)=\frac{123}{35}$

Step 12:Final Answer(part d)

$E\left(X\right)=\frac{123}{35}$