Probability Question

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 your reasoning.

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. Now it is 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) First let's get a sum of all the balls: 10 + 15 + 25+ 30 + 20 = 100 total balls. So, the probability of the first ball being red is 10/100 or 1/10, the probability of the second ball being red is 9/99 or 1/11, and the probability of the third ball being red is 8/98 or 4/49. Multiplying these probabilities together, we get a combined probability of 4/5390 or 2/2695, which in decimal form is 7.42 x 10^-4 and in percent form is 0.0742%.

b) Nope! We already did the three balls for a dime each to calculate our answer in part a), so let's do them all as a group now to show that we get the same result. Suppose we had to group all of the super balls into groups of 3. We'll refer to each super ball by its color and number, which means we have Red1-Red10, Blue1-Blue15, etc. We can't repeat balls, so the total possible groups of 3 is 100 * 99 * 98, which is 970,200 combinations.There are 10 * 9 * 8 or 720 possible combinations of just red balls, which means the probability of choosing 3 reds is 720/970,200, which simplifies to 2/2695 if you divide the numerator and denominator by 360. This is the same probability as in part a), and so we can see that it does not matter whether you do it with a dime or with a quarter.

c) There are other ways to do this one, but I'm going to find the probabilities of the various ways to get this final combination and add them.

Blue->Purple->Green Probability: 15/97 * 30/96 * 20/95

Blue->Green->Purple Probability: 15/97 * 20/96 * 30/95

Purple->Blue->Green Probability: 30/97 * 15/96 * 20/95

Purple->Green->Blue Probability: 30/97 * 20/96 * 15/95

Green->Blue->Purple Probability: 20/97 * 15/96 * 30/95

Green->Purple->Blue Probability: 20/97 * 30/96 * 15/95

We can see that all of these combinations have the same numerators and denominators, just mixed up in different orders. Multiplying them out, all the probabilities listed above are 0.01017, and summing them gives 0.06104 (0.01017 * 6 = 0.06102, not 0.06104, but I kept more decimals places than shown in the answer to avoid rounding error). So, the probability of getting this combination was around 6.1%.

d) It will not make a difference whether you insert a blue, green, or purple ball into the machine; they all affect the probability of you getting a red equally. However, there is a difference between inserting 1 ball and getting 1 ball back vs. inserting 2 of the balls and getting 2 back vs. inserting all 3 balls and getting all 3 back.

1-ball:

Suppose you insert one of the balls back in. This raises the total number of balls left from 94 to 95. There are 7 red balls in the machine (after the person before you miraculously got 3), and so your chances of snagging one will be 7/95 or 0.0737.

2-balls:

It was unclear whether this was an option given that the original purchase had to be in groups of 1 or 3, but let's assume that you can return 2 balls. This would up the total number of balls from 94 to 96. The probability of getting exactly one red back at this point is 7/96 * 89/95 + 89/96 * 7/95 which is 1246/9120 or 0.137.

3 balls:

Suppose you put all the balls back in. This would raise the total from 94 to 97. The probability of getting exactly one red back at this point is 7/97 * 90/96 * 89/95 + 90/97 * 7/96 * 89/95 + 90/97 * 89/96 * 7/95 = 7/97 + 90/97 * 1288/9120 = 168,210/884,640 = 0.190.

Based on the probabilities of getting exactly one red back for each scenario, you should put all 3 balls back into the machine to get the highest chance of getting exactly 1 red, and this chance will be 0.190.

Side Notes for Extra Explaining:

For calculating for the chance of at least one red in the 2 balls case, I found the probability of getting a red on the first try (7/96) and the probability of NOT getting a red on the second try (89/95) and added the probability of NOT getting a red on the first try (89/96) and then getting a red on the second try (7/95).

For calculating the the chance of at least one red in the 3 balls case, I found the probability of the following three combinations:

Red->NOT Red->NOT Red Probability: 7/97 * 90/96 * 89/95

NOT Red->Red->NOT Red Probability: 90/97 * 7/97 * 89/95

NOT Red->NOT Red->Red Probability: 90/97 * 89/96 * 7/95

Adding these gives the probability that exactly 1 ball of the three is red.

Answer 2

a. The number of total balls is now down to 97, since three red balls are gone, so only 7 red balls left. Just do (7/97)*(7/97)*(7/97), since the probability of getting one red ball is 7/97 now.

b. Yes, since you are dealing with three different turns, so paying with a quarter would mean your probability will be 1/1000 as the probability of getting one red ball is 10/100, which is the same as 1/10, so just do (1/10)*(1/10)*(1/10). However, making three separate turns means you must do (10/100) * (9/99) * (8/98), since the total number of balls will decrease by one in each turn along with the number of red balls.

c. You would be down to 97 balls as three of the balls were acquired by the previous person. So probability of getting one green ball is 20/97, blue ball is 15/97, and purple ball is 30/97, so just do (15/97) * (20/97) * (30/97) to get the probability.

d. The probability of getting exactly one red ball would be (10/100) * (90/100) * (90/100). You should replace the balls with a lower probability of obtaining, so that your chances of getting a red ball would be higher.