Question 647490: there are 100 cans of soda.20 cola, 11 orange 29 ginger ale. if one can is selected at random what is the probability that the can will be a rootbeer orange or ginger ale
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! First of all, I think you may have left out information. By the context of the question, I assume there are also 50 cans of root beer.
Let S = the sample space of all cans. In this case, there are 100 elements in the sample space. {20 cola, 11 orange, 29 ginger ale, 50 root beer}
Then the probability of getting root beer or orange or ginger ale, denoted as P(root or orange or ginger) = N(root)+N(orange)+N(ginger)/ N(S). N denotes how many items are in that set.
Let's stop being formal for a minute. Let's just think about if there are 100 cans of soda and you want to pick out root beer. There are 50 cans of root beer, so the probability of picking out root beer is 50/100 or 1/2. This can just be expressed as N(root beer) / N(S) = 50/100.
The question asks us "or". So we need to combine all the probabilities together. There are 50 root beers, 11 orange, and 29 ginger ale. 50+11+29 = 80. That means 80 of the cans that we choose from will satisfy our condition.
So the probability is simply 80/100. This could have been done easier, I know, but I want you to understand what is happening. If you want, reduce 80/100 = 
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