Consider an experiment W that consists of withdrawing a ball from the box. repla
ID: 3132783 • Letter: C
Question
Consider an experiment W that consists of withdrawing a ball from the box. replacing it. and withdrawing a second ball There are red. and I green ball in the box What is the sample space of this experiment? Is this a random variable? Why or why not ' Suppose that the experiment is carried further In counting the number ol red balls ' Why is W now a random variable? For the RV w. find the outcomes as fractions and find the expectation Find the variance for W If X is an RV indicating whether the first ball is red or not (1.0 respectively) and V is whether the second ball is red or not (1.0 respectively). Find Li(X) and E(Y) and show that their sum corresponds to I (W)Explanation / Answer
a.) Sample space: [R,R][R,G][G,R][G,G].
This is not a random variable because we are not calculating probability of particular event.
b.) Here we are calculating aor trying to find out probability of a particular event, i.e. number of red balls in this case, thus W becaomes a random variable.
c.) P(R,R) = 3/4*3/4
= 9/16
P(R,G) = 3/4*1/4
= 3/16
P(G,R) = 1/4*3/4
= 3/16
P(G,G) = 1/4*1/4
= 1/16
E(W) = 2*9/16 + 1*3/16 + 1*3/16 + 0*1/16
= 1.5
d.) Var(W) = E((X-µ)2)
= (2-1.5)2 * 9/16 + (1-1.5)2 * 3/16 + (1-1.5)2 * 3/16 + (0-1.5)2 * 1/16
= 0.375
e.) E(X) = 1*9/16 + 1*3/16 + 0*3/16 + 0*1/16
= 0.75
E(Y) = 1*9/16 + 0*3/16 + 1*3/16 + 0*1/16
= 0.75
E(X) + E(Y) = 1.5 = E(W)
f.) Var(X) = (1-0.75)2*9/16 + (1-0.75)2*3/16 + (0-0.75)2*3/16 + (0-0.75)2*1/16
= 0.1875
Var(Y) = (1-0.75)2*9/16 + (0-0.75)2*3/16 + (-0.75)2*3/16 + (0-0.75)2*1/16
= 0.1875