Bunuel wrote:
When a cookie is taken at random from a jar, what is the probability that it is chocolate flavored?
(1) There are twice as many chocolate flavored cookies as there are almond flavored cookies in the jar.
(2) One third of the cookies in the jar are almond flavored.
Lets assume that there are 3 kind of cookies in the jar :- Chocolate, Almond, Strawberries = C number of total Cookies
(1) There are twice as many chocolate flavored cookies as there are almond flavored cookies in the jar.
Almond = x ; chocolate = 2x ; strawberry = s
x+2x+s=C
Probablity of chocolate = \(\frac{2x}{x+2x+s}\) or \(\frac{2x}{C}\)
Use either equation :- We dont know how many strawberry cookies are there or how many total cookies are there (No info on C or s)
INSUFFICIENT
(2) One third of the cookies in the jar are almond flavored.
This one is even more restrictive and uselss on its own
Almond = \(\frac{1}{3}\)of total C ; chocolate and strawberry = \(\frac{2}{3}\) of total C
Since we dont know the total number of cookies therefore INSUFFICIENT
MERGE BOTH
Almond = \(\frac{1}{3}\) of total cookies C= \(\frac{1C}{3}\)
Chocolate = twice of almond = \(2*\frac{1C}{3}\) of total cookies = \(\frac{2C}{3}\)
BUT WAIT :------> \(\frac{1C}{3} + \frac{2C}{3}= \frac{3C}{3}= C\)which is the number total cookies
SO there is no strawberry cookies ; we only have almond and chocolate cookies ; This proves mathematically that we cannot have Strawberry cookies at all.
Now we are absolutely sure that there are only two types of cookies and thus the two statement can be used without any fear of thinking about any other types of cookies.
So probability of chocolate Cookies = \(\frac{Chocolate}{Total}\)= \(\frac{2C}{3}\)out of C= 2/3
SUFFICIENT
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