Question

5. (15 pts) Let S denote the sample space of tossing the HK dollar coin 9 times with success probability pon the Number side and failure probability g = 1-pon the Flower side. For i=1,2,..., 100, let X, denote the random variable on 2, having value 1 for the outcomes w i th in the number sicle and zero otherwise. Let Y = 3.X1 +3.X2 + ... +3X100- (a)(2 pts) Are the random variables X1,..., X, independent? (b)(3 pts) Find the expectation E(Y) of the random variable y (c)(4 pts) Find the standard deviation (Y) of the random variable y (a) 6 pts) Find the conditional probability PſY = 180 X = ... = X 10 = 0,191 = ... = X100 = 1), i.c., the probability that the Number side appears exactly fifty times, given that the first ten tosses are the Flower sides and the last ten tosses are the Number sides,

Answer 1

a) Yes all these random variables are independent since a toss at any instance doesn't influence the outcome of a toss in another.

b) We will calculate the expected value for any one one $X_i$ and then we will have it for all since the distributions are all the same.

EX = 1.p+0•q =p

So, $E[Y]=E[3 \sum_{i=1}^{100} X_i]=3\sum_{i=1}^{100}E[X_i]=3 \cdot 100 \cdot p=300p$

c) Again the variables are independent so $V[Y]=V[3 \sum_{i=1}^{100} X_i]=9\sum_{i=1}^{100}V[X_i]$ . Also the variance will be the same since they have the same distribution.

EX = 12 ·p+ 02 -q = p

Thus, $V[X]=E[X_i^2]-(E[X_i])^2=p-p^2$

$V[Y]=9 \cdot 100 \cdot (p-p^2)=900(p-p^2)$

So, $\sigma[Y]=\sqrt{V[Y]}=\sqrt{900(p-p^2)}=30\sqrt{p(1-p)}$

d)

$P\left(Y=180|X_1=...=X_{10}=0,X_{91}=...=X_{100}=1 \right ) \\ =P\left(3\sum_{i=1}^{100}X_i =180|X_1=...=X_{10}=0,X_{91}=...=X_{100}=1\right) \\ =P\left( \sum_{i=1}^{10}1+\sum_{i=11}^{90}X_i=60+0+...+0|X_1=...=X_{10}=0,X_{91}=...=X_{100}=1\right ) \\ =P\left( \sum_{i=11}^{90}X_i=50|X_1=...=X_{10}=0,X_{91}=...=X_{100}=1\right )$

Now note that the terms inside the sum $\left ( X_{11},X_{12},...,X_{90} \right )$ are independent of the terms being conditioned on above. Hence, this conditional probability simplifies and we have:

$P\left( \sum_{i=11}^{90}X_i=50|X_1=...=X_{10}=0,X_{91}=...=X_{100}=1\right )=P\left( \sum_{i=11}^{90}X_i=50 \right)$

This is a binomial probability.

For that sum to be 50 we must have exactly 50 tosses where the number side comes up and 80-50=30 times out of the total of 80 tosses. This happens with probability $p^{50}\cdot q^{30}$ . But we have to consider all possible combinations which is $\binom{80}{50}$ . Thus, the probability is $\binom{80}{50} p^{50}\cdot q^{30}$