I need these problems plz

Question

There is a super ball machine in your grocery store that will give you one super ball for a dime and three super balls for a quarter. There are five colors of super balls. This machine is special, because if you don’t like the colors that you get, you can put the super balls back in and try again.

colors Red Blue Black Purple Green

number of each 10 15 25 30 20

a: The person before you got three red balls. What was the probability of that happening? Explain.

b: Is the probability any different if you buy three single super balls for a dime each, or buy three super balls at once for a quarter? Explain.

c: Then it’s your turn at the machine. Remember the person in front of you got three red so the amounts are different than in the chart. You insert a quarter and get blue, purple, and green. What was the probability of that happening? Explain.

d: You really want exactly one red ball and don’t care about the other colors. Which ball(s) should you replace and try again to give the best chance of getting one red ball? If you do this, what is the probability of getting exactly one red ball on your next try?

Answer 1

a) P=1/100+1/100+1/100=3/100=.03=3%

Answer 2

A. The probability of getting 3 separate red super balls for a dime each will have a dependent event if the red ball doesn't get put back into the machine (no replacement). There are a total of 100 super balls (10 reds, 15 blues, 25 blacks, 30 purples, and 20 greens). Let's find the probability of picking the first red, second red, and third red without any particular order and without replacement.

P(3 single red super balls) = (10/100)*(9/99)*(8/98) = 7.42×10^-4 = 0.000742

B. Let's find the probability of picking 3 red super balls at once for a quarter.

P(3 red super balls at once) = C(10,3)/C(100,3) = 120/161,700 = 7.42×10^-4 = 0.000742

The probability of buying 3 single super balls for a dime each is the same as the probability of buying 3 super balls at once for a quarter.

C. The machine will have 97 super balls remaining after the 1st person got 3 red super balls. That also means there are 7 red super balls left in the machine. The next person will get 3 super balls with different colors. The order in this case don't matter.

P(Blue, Purple, Green) = P(Blue)*P(Purple)*P(Green) = (15/97)*(30/96)*(20/95) = 0.01017

This will have the same probability if we have the following possible outcomes:

Blue, Purple, Green → (15/97)*(30/96)*(20/95)

Blue, Green, Purple → (15/97)*(20/96)*(30/95)

Purple, Blue, Green → (30/97)*(15/96)*(20/95)

Purple, Green, Blue → (30/97)*(20/96)*(15/95)

Green, Purple, Blue → (20/97)*(30/96)*(15/95)

Green, Blue, Purple → (20/97)*(15/96)*(30/95)

Therefore, P(Blue, Purple, Green) = 6*(0.01017) = 0.06102

D. This won't matter if you put the blue, purple, or green ball back in the machine. They will affect the probability of getting a red ball equally.

1-ball replacement: If you put one of the balls back in, then the remaining balls will be from 94 to 95. There are 7 red balls in the machine after the previous person got 3 red balls. The probability of getting one red ball is 7/95 = 0.0737.

2-ball replacement: If you put two of the balls back in, then the remaining balls will be from 94 to 96. There are 7 red balls in the machine after the previous person got 3 red balls. The probability of getting one red ball from there is (7/96)*(89/95) + (89/96)*(7/95) = 0.06831 + 0.06831 = 0.13662.

3-ball replacement: If you put all three balls back in, then the remaining balls will be from 94 to 97. There are 7 red balls in the machine after the previous person got 3 red balls. The probability of getting one red ball from there is as follows:

P(First red super ball) = P(RR^cR^c) = (7/97)*(90/96)*(89/95) = 0.06338

P(Second red super ball) = P(R^cRR^c) = (90/97)*(7/96)*(89/95) = 0.06338

P(Third red super ball) = P(R^cR^cR) = (90/97)*(89/96)*(7/95) = 0.06338

P(At least 1 red ball) = 3*(0.06338) = 0.1901

You will have a better chance of getting exactly one red ball if you replace all 3 balls and get the 3 balls back.

Answer 3

There are 100 balls total to start.

a. Probability of 3 reds is 10/100 x 9/99 x 8/98 = 2/2695

b. it makes no difference other than you save a nickel by buying 3 for a quarter

c. 15/97 x 30/96 x 20/95

d. 7/97 x 90/96 x 89/95 for exactly one red ball. replace any non-red ball

multiply the probability of each draw together, the denominator is the total number of balls, which reduces by 1 after each draw. The numerator is the number of ways of getting the desired ball.