SOLUTION: A population can be divided into two subgroups that occur with probabilities 60% and 40%, respectively. An event A occurs 10% of the time in the first subgroup and 50% of the time

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Question 1206322: A population can be divided into two subgroups that occur with probabilities 60% and 40%, respectively. An event A occurs 10% of the time in the first subgroup and 50% of the time in the second subgroup. What is the unconditional probability of the event A, regardless of which subgroup it comes from?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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A population can be divided into two subgroups that occur with probabilities 60% and 40%, respectively.
An event A occurs 10% of the time in the first subgroup and 50% of the time in the second subgroup.
What is the unconditional probability of the event A, regardless of which subgroup it comes from?
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        I changed/edited the condition of the problem
        according to my understanding on how is  SHOULD  BE.

        I firmly believe that without such editing the problem can not be solved. 


The probability that event A will happen in the first group is 0.1*0.6.

The probability that event A will happen in the second group is 0.5*0.4.


Since the groups are , the probability that event A will happen in either group
is the sum

    P = 0.1*0.6 + 0.5*0.4 = 0.06 + 0.2 = 0.26.    ANSWER

Solved.



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Get some graph paper. Draw a square with one vertex at (0,0) and the other at (10,10).
You will have a 10 by 10 square.

Draw a vertical line through x = 6 on the x axis.
The left portion is 6/10 = 60% of the original square.
The right portion is 4/10 = 40% of the original square.

Next, draw a horizontal line through y = 9. Only do so for the left-most region. The smaller rectangle up top represents 10% of the 60% region
This is the blue shaded area in the diagram below.

It is 6 units across and 1 unit tall. The blue area is 6*1 = 6 square units.

Draw another horizontal line through y = 5. Only do this for the right-most region. We split this portion in half.
Shade one of those halves, let's say the bottom half, and that shaded portion is 4 by 5. It has area 4*5 = 20 square units. This area is marked in red.

Add up the shaded areas
6+20 = 26
This is out of a total original area of 10*10 = 100 square units.

26/100 = 13/50 = 0.26 is the probability event A occurs regardless of which subgroup it comes from.
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