SOLUTION: A fair coin is tossed. If a head occurs, 1 die is rolled; if a tail occurs, 2
dice are rolled. Let X be the total on the die or dice. What is the expected
value of X?
Algebra ->
Probability-and-statistics
-> SOLUTION: A fair coin is tossed. If a head occurs, 1 die is rolled; if a tail occurs, 2
dice are rolled. Let X be the total on the die or dice. What is the expected
value of X?
Log On
Question 1149654: A fair coin is tossed. If a head occurs, 1 die is rolled; if a tail occurs, 2
dice are rolled. Let X be the total on the die or dice. What is the expected
value of X? Answer by jim_thompson5910(35256) (Show Source):
In math notation,
---------------------------------------
The probability of landing on heads is 1/2
The probability of rolling any given side of a die is 1/6
The probability of getting heads and getting any given side of a die is (1/2)*(1/6) = 1/12
If the coin lands on heads
X
P(X)
X*P(X)
1
1/12
1*(1/12) = 1/12
2
1/12
2*(1/12) = 2/12
3
1/12
3*(1/12) = 3/12
4
1/12
4*(1/12) = 4/12
5
1/12
5*(1/12) = 5/12
6
1/12
6*(1/12) = 6/12
Sum the values in the third column
1/12 + 2/12 + 3/12 + 4/12 + 5/12 + 6/12 = (1+2+3+4+5+6)/12 = 21/12 = 7/4 = 1.75
We'll use the value 1.75 later on.
---------------------------------------
The probability of landing on tails is 1/2
The chart below represents rolling two dice, one red and one blue, then summing the values. Each sum is shown in black.
Use the chart to see that
probability of rolling a 2 is 1/36
probability of rolling a 3 is 2/36
probability of rolling a 4 is 3/36
probability of rolling a 5 is 4/36
probability of rolling a 6 is 5/36
probability of rolling a 7 is 6/36
probability of rolling a 8 is 5/36
probability of rolling a 9 is 4/36
probability of rolling a 10 is 3/36
probability of rolling a 11 is 2/36
probability of rolling a 12 is 1/36
which helps us construct the table below
X
P(X)
X*P(X)
2
(1/2)*(1/36)=1/72
(2)*(1/72) = 2/72
3
(1/2)*(2/36)=2/72
(3)*(2/72) = 6/72
4
(1/2)*(3/36)=3/72
(4)*(3/72) = 12/72
5
(1/2)*(4/36)=4/72
(5)*(4/72) = 20/72
6
(1/2)*(5/36)=5/72
(6)*(5/72) = 30/72
7
(1/2)*(6/36)=6/72
(7)*(6/72) = 42/72
8
(1/2)*(5/36)=5/72
(8)*(5/72) = 40/72
9
(1/2)*(4/36)=4/72
(9)*(4/72) = 36/72
10
(1/2)*(3/36)=3/72
(10)*(3/72) = 30/72
11
(1/2)*(2/36)=2/72
(11)*(2/72) = 22/72
12
(1/2)*(1/36)=1/72
(12)*(1/72) = 12/72
Sum the values in the third column
2/72 + 6/72 + 12/72 + 20/72 + 30/72 + 42/72 + 40/72 + 36/72 + 30/72 + 22/72 + 12/72 = (2+6+12+20+30+42+40+36+30+22+12)/72 = 252/72 = 7/2 = 3.5
We'll use the value 3.5 later on.
---------------------------------------
Add the two sub-results from each section
1.75 + 3.5 = 5.25
(
