No.3 .HELP NEEDED ASAP

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

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

m+n = P(both are same color)

m+n = 8/63+35/63

m+n = 43/63

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

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

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.

11

step-by-step explanation:

32.50-7.50=25

225/25=11