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Mathematics 16 Online
OpenStudy (anonymous):

In a large bag of 100 Skittles®, each of the 5 colors (red, orange, yellow, blue, green) occurs with the same probability. You reach in and select 2 candies. Find the probability that exactly one of the candies is blue.

OpenStudy (ash2326):

Given that each of the five colors have equal probabilty so each has probability \[P(red)=P(orange)=P(yellow)=P(blue)=P(green)=\frac 15\] Do you get this part

OpenStudy (anonymous):

yeah i under stand the 1/5 but then you pull out two candies which makes it 1/10th right??

OpenStudy (ash2326):

Nope, it's not like that, Let's work step by step

OpenStudy (anonymous):

ookkk

OpenStudy (ash2326):

Do you know combinations?

OpenStudy (anonymous):

yeah the blue is 1/5

OpenStudy (ash2326):

there are 20 blue skittles

OpenStudy (ash2326):

and there are 80 other colored skittles

OpenStudy (anonymous):

oso it would be 1/4 chance of pulling out one blue???

OpenStudy (ash2326):

Nope we have to pull it out of 100, so \[\frac{20}{100}\] Now the other skittle has to be any other color so there are 80 other skittles and we have one skittle already pulled out, so we have 99 skittles \[\frac{80}{99}\] so the probability is \[\frac{20}{100} \times \frac{80}{99}\]

OpenStudy (ash2326):

Do you get this?

OpenStudy (anonymous):

so it would be 16%

OpenStudy (ash2326):

Nope a little part is left, first tell me did you understand till here?

OpenStudy (anonymous):

yeah

OpenStudy (ash2326):

You see we pulled a blue first and then pulled other color, we could have first pulled other color and then blue, in that case tell me what would be the probability?

OpenStudy (anonymous):

i have no clue

OpenStudy (ash2326):

Just read the previous part and try to arrive at the probability

OpenStudy (anonymous):

i still get 16%

OpenStudy (ash2326):

yeah it'd be the same, add the two

OpenStudy (anonymous):

so 18%

OpenStudy (ash2326):

uhuh, \[\frac{20}{100}\times\frac{80}{99}+\frac{80}{100}\times\frac{20}{99}\]

OpenStudy (anonymous):

thanks

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