Question 1163247
<font color=black size=3>
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: <font color=red>1/2</font>


==============================================
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: <font color=red>19/50</font>
</font>