A cup contains 5 red marbles and 4 yellow marbles. Two marbles are removed one at a time replacement what is the probability of selecting a red marble and a yellow marble in either order

Question

a cup contains 5 red marbles and 4 yellow marbles find the probability of selecting a red marble and a yellow marble in either way

Answer 1

5 red 4 yellow

pick 1, replace it, pick a 2nd

P(r)= 5/9

P(y)=4/9

P(r&y)= (5/9)(4/9) =20/81

P(2r)= 25/81

P(2y)= 16/81

P(r&y or y&r)= (81-25-16)/81= 40/81=about 49.38% chance

= about 49%

= almost 50%

but if the marble is not replaced

then

P(r 1st) = 5/9

P(y 2nd) = 4/8

P(r&y) = (5/9)(4/8) = 20/72=5/18

P(y 1st) = (4/9)

P(r 2nd) = 5/8

P(y&r) = (4/9)(5/8) = 20/72= 5/18

P(r&y or y&r) = (5/18)+(5/18) =10/18 = 5/9

=55.555...%

= about 55.56%

= about 56%

without replacement

P(r 1st) = 5/9

P(2r) = (5/9)(4/8) = 5/18

P(y 1st) = 4/9

P(2y) =(4/9)(3/8) = 3/18

P(y&r or r&y) = (18-5-3)/18 = 10/18 =5/9

= about 55.56%

= about 56%