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) ].

Question

Mathematics

Amaya Pineda

3 May, 20:26

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) ].

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Marisol · Answer 1

a) the joint probability distribution function is

P (y1, y2) =

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) the probability is 50% (F (2,1) = 1/2)

Step-by-step explanation:

for the variable Y1, we start with the negative binomial distribution, with true or false values:

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