Part A
2 white, 5 red, 7 total
P(1st white) = 2/7
P(1st white, 2nd white) = P(1st white)*P(2nd white)
P(1st white, 2nd white) = (2/7)*(4/9)
P(1st white, 2nd white) = 8/63
Let m = 8/63
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P(1st red) = 5/7
P(1st red, 2nd red) = P(1st red)*P(2nd red)
P(1st red, 2nd red) = (5/7)*(7/9)
P(1st red, 2nd red) = 5/9
Let n = 5/9 = 35/63
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m+n = P(both are same color)
m+n = 8/63+35/63
m+n = 43/63
Answer: 43/63
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Part B
m = P(first white then red)
m = (2/7)*(5/9)
m = 10/63
n = P(first red then white)
n = (5/7)*(2/9)
n = 10/63
r = P(second bead isn't same color as first)
r = m+n
r = 10/63+10/63
r = 20/63
Answer: 20/63
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Part C
m = P(1st white, 2nd white)
m = (2/7)*(4/9)
m = 8/63
n = P(1st red, 2nd white)
m = (5/7)*(2/9)
m = 10/63
m+n = P(2nd is white)
m+n = 8/63+10/63
m+n = 18/63
m+n = 2/7
Answer: 2/7
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Part D
If we withdraw 1 white bead, but we add it back in, then the probability of selecting white has not changed on the second selection.
The same can be said if we picked a red bead as well. This is because we simply replace the red bead with a copy of some other red bead.
Nick's statement is another way of saying replacement is happening.
Answer: True