SOLUTION: A bowl contains orange and green candles. The probability of randomly selecting an orange can is 1/4 and the probability of randomly selecting a green candy is 1/6. Which of the fo

Algebra ->  Statistics  -> Normal-probability -> SOLUTION: A bowl contains orange and green candles. The probability of randomly selecting an orange can is 1/4 and the probability of randomly selecting a green candy is 1/6. Which of the fo      Log On


   



Question 1141566: A bowl contains orange and green candles. The probability of randomly selecting an orange can is 1/4 and the probability of randomly selecting a green candy is 1/6. Which of the following could be the total number of candles in the bowl.
A)10
B)12
C)18
D)30

Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Want the common denominator of 4 and 6. That would be 12 (or 24, 36, 48).
B

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = # of orange candles and y = # of green candles.


Then from the condition


x%2F%28x%2By%29 = 1%2F4,      (1)

y%2F%28x%2By%29 = 1%2F6.      (2)


From this system


4x = x + y                (3)

6y = x + y                (4)


To solve it, express  y = 4x - x = 3x  from equation (3) and substitute it into equation (4). You will get


6*(3x) = x + 3x,

18x = 4x


It has the solution  x= 0  ONLY.


Similarly,  y= 0 is the only solution for y.


ANSWER.  The condition describes an IMPOSSIBLE situation.


         The events, described in the post, NEVER MAY happen.

Solved.

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An auxiliary note :

    The condition mixes candles, cans and candies by a bizarre way.