Problem with percentages

**jlee** · July 9th 2008, 07:50 AM

I am having trouble with the following problem, your help is greatly appreciated.

Super Cola sales breakdown as 80% regular soda and 20% diet soda. While 60% of the regular soda is purchased by men, only 30% of the diet soda is. If a woman purchases Super Cola, what is the probability that it is a diet soda? (Hint: The 60% and 30% values are conditional probabilities stated as percentages.)

**colby2152** · July 9th 2008, 08:26 AM

Originally Posted by jlee

I am having trouble with the following problem, your help is greatly appreciated.

Super Cola sales breakdown as 80% regular soda and 20% diet soda. While 60% of the regular soda is purchased by men, only 30% of the diet soda is. If a woman purchases Super Cola, what is the probability that it is a diet soda? (Hint: The 60% and 30% values are conditional probabilities stated as percentages.)

You want to find:

$P(Diet | Woman)$

Conditional Probability states that

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

So...

$P(Diet | Woman) = \frac{P(Diet \cap Woman)}{P(Woman)}$

$P(Diet \cap Woman) = 0.2* 0.7 \Rightarrow 0.14$ *

*Why? Well, 70% of diet pop is consumed by women, but diet pop is only 20% of Super Cola sales, so as a whole, the percentage of Super Cola customers who are women and drink diet soda is 14%

$P(Woman) = ???$

Well, the amount of men buying Super Cola is a weighted sum:

$P(Man) = 0.6*0.8 + 0.3*0.2 \Rightarrow 0.48 + 0.06 = 0.54$

$P(Woman) = 1 - P(Man) \Rightarrow 0.46$

--------------------

$P(Diet | Woman) = \frac{P(Diet \cap Woman)}{P(Woman)}$

$P(Diet | Woman) = \frac{0.14}{0.46}$

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