Study of Total Probability is as follows
Let a1, a2 and a3 be mutually exclusive and exhaustive events and let b indicate some other event. Note for a to arise it has to be in conjunction with at least one of a1, a2 and a3. Therefore, by Submitting the multiplication rule to a general situation of this type we obtain the following.
The Law of Total Probability
Let a1, a2, . . . ,an be a mutually exclusive and exhaustive events. If B is any other event it follows that
P(B) = P(B|A1)+P(B|A2)+…+P(B|AN) = `sum_(j=1)^n P(B|Aj)`
When studying total probability this formula have to be used.
Examples for Total Probability
Example 1:
A fair coin is tossed. If the coin lands on heads a bag is filled with one black ball and three white balls. If the coin landed on tails the bag is filled with one black ball and nine white balls. A ball is then selected from the bag. What is the probability that the ball selected is black?(Use Total Probability)
Solution
Let H = Heads, T = Tails and B = Black ball selected. Then by the law of total probability
P(B) = P(B|H)P(H) + P(B|T)P(T)
= `(0.25)(0.5) + (0.1)(0.5)`
P (B) = 0.175
Example 2:
Three boxes contain red and green balls. Box 1 has 5 red balls and 5 green balls, Box 2 has 7 red balls and 3 green balls and Box 3 contains 6 red balls and 4 green balls. The probabilities of choosing a box are 1/4, 1/6, 1/8. What is the probability that the ball chosen is green by using the Total Probability?
Solution
We begin by defining the following sets. Let,
G:= the ball chosen is green.
B1:= Box 1 is selected
B2:= Box 2 is selected
B3:= Box 3 is selected
Then P(G|B1) = 5/10, P(G|B2) = 3/10 and P(G|B3) = 4/10.
P(G) =` ((5/10)xx(1/4)) + ((3/10)xx(1/6))+((4/10)xx(1/8))` = 9/(40)
Example 3:
Three shed car colors red and green. Shed 1 has 10 cars of red and 10 cars of green, shed 2 has 7 cars red and 3 green cars and shed 3 contains 6 red cars and 4 green cars. The respective probabilities of choosing a shed are 1/4, 1/6, 1/8. Use the Total Probability to find the probability that the car chosen is green?
Solution
We begin by defining the following sets. Let,
G = the ball chosen is green.
S1 = Shed 1 is selected
S2 = Shed 2 is selected
S3 = Shed 3 is selected
Then P(G|S1) = 10/20, P(G|S2) = 3/10 and P(G|S3) = 4/10.
P(G) =` ((10/20)xx(1/4)) + ((3/10)xx(1/6))+((4/10)xx(1/8))` = 9/(40)