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Topic: Union of probabilities
Replies: 2   Last Post: Jan 12, 2010 7:23 PM

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Nagesh

Posts: 2
From: Bangalore, INDIA
Registered: 1/12/10
Union of probabilities
Posted: Jan 12, 2010 4:09 AM
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Consider a box B1 that has R1, G1 and B1 number of red, green and blue balls. There is another box B0 which will hold balls drawn from B1. There are only r1, g1 and b1 number of red, green and blue balls out of R1, G1 and B1 that are eligible to be present in B0. There are similar boxes B2 and B3 with its own count of red, green and blue
balls and its own count of eligible red, green and blue balls. What is the probability of picking up at least one eligible red ball after drawing balls from B1, B2 and B3 ?

My solution attempt :
The probability of picking up a eligible red ball (from B1) - conditional on drawing a red ball - would be r1/R1. From B2, it would be r2/B2 and from B3, it would be r3/R3. These events are independent but not mutually exlcusive so we have to take a union of the probabilities from three boxes.

P(B1) = r1/R1, P(B2) = r2/R2, P(B3) = r3/R3
P(at least one eligible red ball from three boxes) = P(B1 U B2 U B3)
= P(B1) + P(B2) + P(B3) - P(B1 int B2) - P(B2 int B3) - P(B1
int B3) + P(B1 int B2 int B3)
= r1/R1 + r2/R2 + r3/R3 - r1+r2)/(R1+G1+B1+R2+B2+G2) - (r2+r3)/(R2+G2+B2+R3+G3+B3) - (r1+r3)/(R1+G1+B1+R3+G3+B3) + (r1+r2+r3)/(R1+G1+B1+R3+G3+B3+R2+B2+G2)

int => intersection

Am I doing the union of probabilities correctly ?



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