Consider a random experiment that begins by flipping four coins: two blue coins
ID: 3302791 • Letter: C
Question
Consider a random experiment that begins by flipping four coins: two blue coins and two green coins. If blue coins show exactly one head, then define random variable Xas the total number of heads shown on the four coins; otherwise, define X as the number of tails shown by all of the coins. The blue coins are fair, while the green coins are weighted to show heads with probability 2/3. All coins are independent. (a) Determine the probability mass function for A. (b) Suppose that your friend completes this experiment and tells you that their observed value of X was three. What is the probability that both green coins showed heads? [HINT: Use the definition of conditional probability]Explanation / Answer
a)
P(X=0) =P( both heads on blue coins *both heads on green coins)
=(1/4)*(2/3)*(2/3) =4/36
P(X=1) =P( one head on blue coins* no head on green coins+both head on blue coin and one tail on green coin)
=(1/2)*(1/3)*(1/3)+(1/4)*((2/3)*(1/3)+(1/3)*(2/3)) =6/36
P(X=2) =P( one head on blue coins* one head on green coins+both head on blue coins* both tails on green +both tails on blue*both heads on green)
=(1/2)*((1/3)*(2/3)+(2/3)*(1/3))+(1/4)*(1/3)*(1/3)+(1/4)*(2/3)*(2/3) =13/36
P(X=3) =
P( one head on blue coins* two head on green coins+two tails on blue and 1 tail on green)
=(1/2)*(2/3)*(2/3)+(1/4)*((1/3)*(2/3)+(2/3)*(1/3)) =12/36
P(X=4) =P( both tails on blue and both tails on green) =(1/4)*(1/3)*(!/3) =1/36
therefore below is pmf in table form:
b) from above conditional probability that both green coins shows heads given observed value is 3
= P( one head on blue coins* two head on green coins)/P(X=3) ==((1/2)*(2/3)*(2/3))/(12/36)
=8/12=2/3
x P(x) 0 4/36 1 6/36 2 13/36 3 12/36 4 1/36