A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12
My method #1: probability:
# of ways of getting 3 blue hats (w/replacement): 1/2*1/2*1/2*1/2 (the last 1/2 designates the chances of getting a red hat for the fourth selection.) = 1/16
# of ways of getting 3 red hats (w/replacement): 1/2*1/2*1/2*1/2 (the last one-half designates the chances of getting a red hat for the fourth selection.) = 1/16
My answer: 1/8
My Method #2: Combinations
4C1*4C1*4C1*4C1 = 128 (ways to select 3 blue hats)
4C1*4C1*4C1*4C1 = 128 (ways to select 3 red hats)
8C1*8C1*8C1*8C1=4096 (number of ways to pick any 4 hats)
Answer: 128/4096+128/4096
=1/8
Why are they both wrong? The OA is 1/2
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12
My method #1: probability:
# of ways of getting 3 blue hats (w/replacement): 1/2*1/2*1/2*1/2 (the last 1/2 designates the chances of getting a red hat for the fourth selection.) = 1/16
# of ways of getting 3 red hats (w/replacement): 1/2*1/2*1/2*1/2 (the last one-half designates the chances of getting a red hat for the fourth selection.) = 1/16
My answer: 1/8
My Method #2: Combinations
4C1*4C1*4C1*4C1 = 128 (ways to select 3 blue hats)
4C1*4C1*4C1*4C1 = 128 (ways to select 3 red hats)
8C1*8C1*8C1*8C1=4096 (number of ways to pick any 4 hats)
Answer: 128/4096+128/4096
=1/8
Why are they both wrong? The OA is 1/2
