Probability (expectation)

Question

Suppose there are 20 different types of coupons and suppose that each time one obtain a coupon is equal likely to be any one of the type compute the expected number of different types that are contained in a set of 10 coupons?

Answer 1

Blockquote

$20^{10}$ = The number of different ways 10 coupons can be selected from 20 types, assuming repetition is allowed.

The number of ways all 10 coupons are same = $^{20}C_1 \times ^{19} C_0$
The number of ways 9 coupons are same = $^{20}C_1 \times ^{19} C_1$
The number of ways 8 coupons are same = $^{20}C_1 \times ^{19} C _2$
The number of ways 7 coupons are same = ${20}C_1 \times ^{19} C_3$
. . . The number of ways 0 coupons are same = $^{20} C_{10}$

$N = ^{20}C_1 \times ^{19}C_0 + ^{20}C_1 \times ^{19}C_1 + ^{20}C_1 \times ^{19}C_2 + ... + ^{20}C_1 \times ^{19}C_9 + ^{20}C_{10} = 20^{10}$

X = The number of coupons that are different.

$E(X) = \frac{^{20} C_1 \times ^{19}C_0 \times 0 + ^{20}C_1 \times ^{19}C_1 \times 1 + ... + ^{20}C_1 \times ^{19} C_9 \times 9 + ^{20} C_{10} \times 10}{20^{10}}$

EDIT 1 START

It seems my answer is incorrect. I must've misunderstood the question.

I've calculated the expected value for random variable X = the number of different coupons in the 10 selected. Is the answer correct for this interpretation of the question?

EDIT 1 END

Answer 2

Each type has an equal probability of being picked/not picked in each draw, thus

P(type i not picked in a draw) $=\frac{19}{20}=0.95$

so P( type i not picked in $10$ draws) $= 0.95^{10}$

and P(type i picked in 10 draws)$=1 - 0.95^{10}$

By linearity of expectation, expected number of types collected

$$ =\sum_{i=1}^{20}(1- 0.95^{10})\approx 8$$