# I have a bag of candy  with 9 red sweets, 15 orange, and 8 green. If 2 sweets are eaten one after another , what is the probability that the first  sweet eaten was orange and the second one was...

I have a bag of candy  with 9 red sweets, 15 orange, and 8 green. If 2 sweets are eaten one after another , what is the probability that the first  sweet eaten was orange and the second one was red?   Also what is the probability that both sweets are green?

ishpiro | Certified Educator

The probability of getting an orange sweet out of the bag is the ratio of the number of orange sweets to the total number of sweets.

The total number of sweets is 9 + 15 + 8 = 32. There are 15 orange sweets, so the probability of getting an orange sweet is `15/32` .

Since the orange sweet was eaten, it was not replaced in the bag. So now the total number of sweets in the bag is 31. Since there are still 9 red sweets left, the probability of getting a red sweet after an orange one is eaten is `9/31`  .

The probability of getting an orange sweet first and a red sweet second is

`15/32* 9/31 = 135/992` . Converting into percents, this will be about 13.61%

The probability that the first sweet eaten is orange and the second one is red is

135/992, or approximately 13.61%.

To find the probability of getting two green sweets, consider again pulling out a green sweet out of the bag of 32 sweets. Since there are 8 green sweets in the bag, this probability is `8/32 = 1/4` .

Since the first green sweet is not replaced, there are now 7 green sweets in the bag, and the total number of sweets is 31. So the probability of getting a second green sweet is `7/31` .

The probability of getting two green sweets is then

`1/4 * 7/31 = 7/124`

Converting to percents, this will be about 5.65%.

The probability that both sweets are green is 7/124, or 5.65%.

` `

nisarg | Student

first one  being orange is 15/32 red one is then 9/31 both green is 8/32*7/31 so about 7/124

steamgirl | Student

To find the probability  of any of the candies being chosen, you must first know how many are in a bag, and the candy ratios. In this case there are a total of 32 pieces of candy, red had a ratio of 9/32, orange is 15/32 and green is 8/32.

Probability of the first candy being chosen and the ratio above would be the same, so if the first candy to be pulled was orange, then the probability would be 15/32 or 47% (15 divided by 32 times 100 = 47%), and if green were pulled first it would be 8/32 or 25%.

The second candy would be different though, because there are no longer 32 candies in the bag, there are 31. So if the second candy was red, then there would be a 9/31 chance, or 29%, that it would be pulled, or 8/31, 26%, if it were green.

crystaltu001 | Student

The probability that the first one eaten is 15/32. 32 is the total of all the candies added together. The probability that the second one eaten is  9/31.

The probability that both of the sweets that you end up picking out is green is 8/32

rachellopez | Student

Probability is just a part over whole ratio. Since there are 9 red sweets, 15 orange sweets, and 8 green sweets, there are a total of 32 sweets.

The probability that the first sweet would be orange is 15/32 since there are 15 orange sweets out of 32 total.

The probability of the second sweet being red would be 9/31. The total goes down by one because the orange sweet from part one was not added back to the mix.

For the probability of both being green you have to look at them separately. The probability that the first one is green would be 8/32 which simplifies to 1/4. Since this green sweet was not put back there would only be 7 green sweets left and only 31 total, so this probability would be 7/31. Multiply these together add your probability of both being green is 7/124.