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09 May 2017, 02:53
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68% (01:43) correct
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When a bowl containing 6 black, red or green spheres has 3 spheres randomly drawn from it and replaced the probability of drawing three black spheres is 1/27. If there are x black spheres in the bowl then what is the value of x?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
12 May 2017, 13:31
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Bunuel
We are given that there are 6 black, red, and green spheres in a bowl and 3 spheres are randomly drawn from it with replacement, and that the probability of drawing three black spheres is 1/27. We are also given that there are x black spheres in the bowl and we need to determine the value of x, i.e., the number of black spheres in the bowl.
Since there are x black spheres in the bowl, the probability of drawing a black sphere is x/6. Thus, the probability of drawing 3 black spheres, with replacement, is (x/6)^3. We are given that this probability is 1/27; thus, we have:
(x/6)^3 = 1/27
(x/6)^3 = (1/3)^3
x/6 = 1/3
3x = 6
x = 2
General Discussion
09 May 2017, 06:32
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1/27= 1/3*3*3
Total Sphere: 6
Probabilty:
x/6*x/6*x/6
Only option which satisfy the above equation is x=2
Total Sphere: 6
Probabilty:
x/6*x/6*x/6
Only option which satisfy the above equation is x=2
