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

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

$E\left(X\right)={鈭�}_0^{\mathrm{鈭�}}鈥�\left(1鈭�\underset{i=1}{\overset{4}{鈭�}}鈥�\left(1鈭�e^{鈭�p_{i}x}\right)\right)dx$

we have that

$\underset{i=1}{\overset{4}{鈭�}}鈥�\left(1鈭�e^{鈭�p_{i}x}\right)={\left(1鈭�e^{鈭�x^2/8}\right)}^2{\left(1鈭�e^{鈭�3x^2/8}\right)}^2$

role="math" localid="1647442465837" $=e^{鈭�x^2}鈭�2e^{鈭�\left(7x^2\right)/8}+e^{鈭�\left(3x^2\right)/4}鈭�2e^{鈭�\left(5x^2\right)/8}+4e^{鈭�x^2/2}鈭�2e^{鈭�\left(3x^2\right)/8}+e^{鈭�x^2/4}鈭�2e^{鈭�x^2/8}+1$

Integrate that over the positive real numbers to get that

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

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

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.

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

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

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

Given that all the types of the second group.

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

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

Given that all the types of either group.

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

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