Three balanced coins are tossed independently. One of the variables of interest is Y1, the number of heads. Let Y2 denote the amount of money won on a side bet in the following manner. If the first head occurs on the first toss, you win $1. If the first head occurs on toss 2 or on toss 3 you win $2 or $3, respectively. If no heads appear, you lose $1 (that is, win - $1)a. Find the joint probability function for Y1 and Y2.b. What is the probability that fewer than three heads will occur and you will win $1 or less? [That is find F(2,1)].

p(r,x)=C(x+r-1,r-1)*p^r*(1-p)^x , where x= number of false experiments until a number r of true values is achieved, and C(a,b)=number of combinations of b in a values. if we set r=1 (fist head) and y1=x+1 (except of y1=-1)

p(y1)= C(y1,0)*p^0*(1-p)^(y1-1) = p*(1-p)^(y1-1)

for the variable Y2, we begin with the binomial distribution:

p(n,x)= C(n,x)*p^x*(1-p)^(n-x) , n= number of times the coin is tossed, x= number of heads obtained , p = probability of obtaining heads when the coin is tossed once.

if we set n=3-y1 (remaining coins that can be tail or head), and x=y2-1 (since we already know the first head appeared), we calculate the probability of p(y2) in case of y1 ( or p(y2|y1) )

p(y2|y1)= C(3-y1, y2-1)*p^(y2-1)*(1-p)^(4-y1-y2)

Now using the theorem of Bayes:

P(y2,y1)=p(y2∩y1)=p(y2|y1)*p(y1) = C(3-y1,y2-1)*p^y2*(1-p)^(3-y2)

since the coins are balanced p= 0,5

P(y2,y1)= C(3-y1,y2-1)*0,5^y2*(1-0,5)^(3-y2)= C(3-y1,y2-1)*0,5^3=C(3-y1,y2-1)/8

therefore

P(y2,y1)=

C(3-y1,y2-1)/8, for 1≤y2≤4-y1 and 1≤y1≤3 ,

1/8 for (y1,y2)=(-1 ,0)

0 otherwise

b) F(2,1)= P(y=1,y2<2)+P(y=(-1),y2=0)= [C(2,0)/8+C(2,1)/8] + 1/8 = (1/8+2/8)+1/8=4/8=1/2