Question 1030391: In a bag, there are 5 green candies, 3 red candies, and 7 orange candies. Once a candy is selected, it is not replaced. Find each probability.
P(a red candy and then an orange candy)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 5 green candies
3 red candies
7 orange candies
5+3+7 = 15 candies total
P(red candy) = (# of red)/(# total)
P(red candy) = 3/15
P(red candy) = 1/5
Don't forget to reduce any fractions fully to make the problem simpler
After the first candy is chosen, it is NOT replaced. So we have 15-1 = 14 candies left over
P(orange candy after first candy is picked) = (# of orange candies)/(amount of candy left over)
P(orange candy after first candy is picked) = 7/14
P(orange candy after first candy is picked) = 1/2
Again, don't forget to reduce
So we have these two probabilities 1/5 and 1/2. We now multiply them
P(red then orange) = P(red)*P(orange after first candy is picked)
P(red then orange) = (1/5)*(1/2)
P(red then orange) = (1*1)/(5*2)
P(red then orange) = 1/10
which is our final answer as a fraction. If you wish to report the answer in decimal form, then use a calculator to get 1/10 = 0.1 which converts to 10%. The three all say the same basic thing.
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Final answer as a fraction: 1/10
Final answer as a decimal value: 0.1
Final answer as a percentage: 10%
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