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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]
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p(yellow)=1-p(remaining)
=1-(1/8+1/5+1/10)
=23/40
no of yellow balls = 23
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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]
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Bunuel wrote:
In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

A. 25.
B. 24.
C. 23.
D. 20.
E. 17.


We see that the total number of balls must be a multiple of the LCM of 8, 5, and 10, which is 40. However, since there are no more than 50 balls in the jar, the number of balls must be 40.

Let’s determine the probability of selecting a yellow ball. Since the selected ball will be either blue, red, green, or yellow, we have:

P(blue) + P(red) + P(green) + P(yellow) = 1

1/8 + 1/5 + 1/10 + P(yellow) = 1

5/40 + 8/40 + 4/40 + P(yellow) = 1

17/40 + P(yellow) = 1

P(yellow) = 1 - 17/40 = 23/40

Since the probability of selecting a yellow ball is 23/40, we can create the following proportion to determine the number of yellow balls:

23/40 = y/40

23 = y

Alternative solution:

After we have determined that the total number of balls in the jar must be 40, we can say that the number of blue balls must be 1/8 x 40 = 5, the number of red balls must be 1/5 x 40 = 8, and the number of green balls must be 1/10 x 40 = 4. Thus, the number of yellow balls in the jar must be 40 - (5 + 8 + 4) = 40 - 17 = 23.

Answer: C
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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]
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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]
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