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probabilty problem

Problem Solving — algebra and arithmetic
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by sukh » Fri Jul 15, 2011 7:59 am
A box has 60 red, green, blue and white pieces of candy inside. If one candy is drawn from the box at random and the probability that it will be white is 1/5 and the probability that it will be green is 1/4 , what is the probability that the candy piece will be either red or blue
sorry for posting too many questions
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by vineeshp » Fri Jul 15, 2011 8:21 am
Probability it is white is 1/5 or 12/60
Probability is is green is 1/4 or 15/60.

So there are 27 white and green candies.

so there are 33 red or white candy. so probability is 33/60.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by MBA.Aspirant » Fri Jul 15, 2011 8:26 am
sukh wrote:A box has 60 red, green, blue and white pieces of candy inside. If one candy is drawn from the box at random and the probability that it will be white is 1/5 and the probability that it will be green is 1/4 , what is the probability that the candy piece will be either red or blue
sorry for posting too many questions
prob of white is 1/5 implies there're 60/5 = 12 white

prob of green = 1/4 implies 15 green

so prob red or blue = the rest/total = 33/60 = 11/20
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by sukh » Fri Jul 15, 2011 8:33 am
i solved it by 1- probability of a or b complementary rule
1- 1/4-1/5

another problem The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?
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by sandeeptalla450 » Fri Jul 15, 2011 11:45 am
sukh wrote:i solved it by 1- probability of a or b complementary rule
1- 1/4-1/5

another problem The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?

Probability that he'll not buy it at ANY of the 3 stores = (1/2)*(1/2)*(1/2)=1/8
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