Question 1163247: A bag contains 2 white,5 red and 3 blue balls. One ball is drawn, its color noted and put back in the bag. Then a second ball is drawn. Find the probability that
i) the second ball drawn is either white or blue?
ii) the two balls are of the same color?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Part i)
The first ball is selected and then put back in the bag. This means that the second selection has the same conditions as the first selection. Nothing has changed. So the two selections are independent.
We have 2+5+3 = 10 balls total, of which 2+3 = 5 are white or blue.
The probability of selecting either a white or blue ball is 5/10 = 1/2.
Answer: 1/2
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Part ii)
P(white) = (2 white)/(10 total) = 1/5
A = P(2 white)
A = P(white)*P(white)
A = (1/5)*(1/5)
A = 1/25
Note that P(white) stays the same due to the events being independent (only because we put the first ball back)
P(red) = (5 red)/(10 total) = 1/2
B = P(2 red)
B = P(red)*P(red)
B = (1/2)*(1/2)
B = 1/4
P(blue) = (3 blue)/(10 total) = 3/10
C = P(2 blue)
C = P(blue)*P(blue)
C = (3/10)*(3/10)
C = 9/100
Add the values of A,B,C
Such addition is valid because events A,B,C are mutually exclusive (we can get one or the other, but not multiple at the same time)
A+B+C = 1/25 + 1/4 + 9/100
A+B+C = 4/100 + 25/100 + 9/100
A+B+C = (4+25+9)/100
A+B+C = 38/100
A+B+C = (19*2)/(50*2)
A+B+C = 19/50
The probability of picking either
A) two white
B) two red, or,
C) two blue
is 19/50
Answer: 19/50
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