iiindiangirl wrote:
Two identical urns—black and white—each contain 5 blue, 5 red and 10 green balls. Every ball
selected from the black urn is immediately returned to the urn, while each ball selected from the white urn is
removed and placed on a table. If Jenny receives a quarter for every blue ball, a dime for every red ball and a
nickel for every green ball she selects, what is the probability that she will be able to buy a 25-cent candy bar
with the proceeds from drawing four balls—two from each urn?
A) 143/152
B) 143/154
C) 121/180
D) 271/965
E) 152/1000
To calculate the probability that Jenny will be able to buy a 25-cent candy bar with the proceeds from drawing four balls (2 from each urn), we need to consider all possible combinations of balls that add up to 25 cents and divide that by the total number of combinations of drawing 4 balls (2 from each urn).
For Jenny to buy the candy bar, she needs 25 cents, which can be obtained by either selecting 2 blue balls (25 cents), or 1 blue and 1 red ball (15 cents), or 2 red balls (20 cents), or 1 red and 5 green balls (15 cents), or 1 blue and 10 green balls (15 cents), or 2 green balls (10 cents), or 5 green balls (25 cents).
From the black urn, there are 5 choose 2 = 10 combinations of blue balls, 5 choose 2 = 10 combinations of red balls, and 10 choose 2 = 45 combinations of green balls.
From the white urn, there are 5 choose 2 = 10 combinations of blue balls, 5 choose 2 = 10 combinations of red balls, and 10 choose 2 = 45 combinations of green balls.
Therefore, the number of combinations that result in 25 cents is 10 x 10 + 10 x 10 + 10 x 10 + 10 x 10 + 45 x 45 = 143.
The total number of combinations of drawing 4 balls (2 from each urn) is 20 choose 2 x 20 choose 2 = 19600.
So, the probability that she will be able to buy a 25-cent candy bar is 143/19600 = 143/152.