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.

to determine the unit cost your would divide the total cost by the number sold.

7 bagels for $4.41       $4.41/7 = $.63 each at fresh side

5 bagels for $3.60         $3.60/5 = $.72 each at value mart

its 50 miles away from the starting point