Veronica wrote:A drawer holds 4 red hats and 4 blue hats. What is the probability f 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.1/4; c.1/2; d.3/8; e.7/12
I have read the explanation in the answer key already, but I cannot understand thoroughly. Please help me analyze and solve this problem. Thank you!
Since there are only two options, red and blue hats, and there's an equal number of the two, we know the following:
the probability of selecting a red hat =
1/2
the probability of selecting a blue hat =
1/2
Hats are selected one by one
with replacement
So we have two cases:
Case 1: 3 out of 4 hats are Red
The number of ways we can pick three red hats: (I just listed them)
RED RED RED BLUE
RED RED BLUE RED
RED BLUE RED RED
BLUE RED RED RED
Since each event is independent of the last (the hats are replaced after each selection) we know that selecting either a blue or a red hat will always have the probably of 1/2
Therefore:
RED RED RED BLUE = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
RED RED BLUE RED = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
RED BLUE RED RED = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
BLUE RED RED RED = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
Total for Case 1 = 4/16 = 1/4
Case 2: 3 out of 4 hats are Blue (Same process as case one)
BLUE BLUE BLUE RED = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
BLUE BLUE RED BLUE = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
BLUE RED BLUE BLUE = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
RED BLUE BLUE BLUE = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
Total for Case 2 = 4/16 = 1/4
Probability of Case 1 or Case 2 = Probability of Case 1 + Probability of Case 2 = 1/4 + 1/4 =
1/2
Pick C
