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In a box there are 6 orange balls, 4 green balls and "x" blue balls. If the probability of drawing a blue ball is 1/3, what is the probability of drawing two blue balls?

Sagot :

Step-by-step explanation:

I assume it is meant to draw 2 blue balls in a row out of the full pool of balls (no other draws happen before or during this process).

tricky, your teacher.

I think the temptation is to simply say 1/3×1/3 = 1/9.

but is it ?

the probabilty to draw one blue ball out of the complete system (no ball missing) is

x / (6+4+x) = 1/3

x = (6+4+x)/3

3x = 6 + 4 + x = 10 + x

2x = 10

x = 5

so, we have a system of 6 orange, 4 green and 5 blue balls (15 all together) .

the probability at our first pull to draw a blue ball is the indicated 5/15 = 1/3.

but now, for the second ball, we have only 6 orange, 4 green and 4 blue balls in the pool (14 altogether).

and so the probability to draw a blue ball out of this is

4/14 = 2/7.

therefore, the probability to draw 2 blue balls is now the combination

1/3 × 2/7 = 2/21

aha ! so, it is a little bit less than the originally suspected probability of 1/9.

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