Total Probability :

Total Probability Intro:

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)

 

Examples for Total Probability

Example 1:

A reasonable 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?

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 packs contain red and green balls. pack 1 has 5 red balls and 5 green balls, Pack 2 has 7 red balls and 3 green balls and Pack 3 contains 6 red balls and 4 green balls. The probabilities of choosing a pack are 1/4, 1/6, 1/8. What is the probability that the ball chosen is green?

Solution
we begin by defining the following sets. Let,

G:= the ball chosen is green.
B1:= Pack 1 is selected
B2:= Pack 2 is selected
B3:= Pack 3 is selected

Using Law of total probability

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 hut car colors red and green. Hut 1 has 10 cars of red and 10 cars of green, hut 2 has 7 cars red and 3 green cars and hut 3 contains 6 red cars and 4 green cars. The respective probabilities of choosing a hut are 1/4, 1/6, 1/8. What is the probability that the car chosen is green?

Solution

We begin by defining the following sets. Let,

G = the ball chosen is green.
S1 = Hut 1 is selected
S2 = Hut 2 is selected
S3 = Hut 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)

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