Question 252430
Let N = {1, 2, 3, 4, 5}

(a) without replacement - P(even and even) = 2/5 * 1/4 = 2/20 = 1/10.
A = {(2,4), (4,2)}

(a) with replacement - P(even and even) = 2/5 * 2/5 = 4/25
A = {(2,2), (2,4), (4,2), (4,4)}

(b) without replacement P(odd and odd) = 3/5 * 2/4 = 6/20 = 3/10
B = {(1,3), (1,5), (3,1), (3,5), (5,1), (5,3)}
(b) with replacement P(odd and odd) = 3/5 * 3/5 = 9/25
B = {(1,1), (1,3), (1,5), (3,1), (3,3) (3,5), (5,1), (5,3), (5,5)}

(c) with replacement only: P (X and X) = 1/5 * 1/5 = 1/25
C = {(1,1), (2,2), (3,3), (4,4), (5,5)} - -> only one choice can be used.

(d) without replacement - P(odd and even). This means P(odd and even) OR P(even and odd). So, we have 2 x P(even and odd).

without replacement - 2 x P(even and odd) = 2 * 2/5 * 3/4 = 12/20 = 3/5
D = {(2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (1,2), (1,4), (3,2), (3,4), (5,2), (5,4)}

with replacement - 2 x P(even and odd) = 2 * 2/5 * 3/5 = 12/25.
D = {(2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (1,2), (1,4), (3,2), (3,4), (5,2), (5,4)}

Hope that helps.