A bag contains 5 red marbles, 4 blue marbles and 7 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be green?
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Answer:
There are 12 marbles (5 red, 4 blue and 3 green) in the basket from which 3 marbles can be drawn in C(12,3)=220 ways. Let A be the event that all 3 marbles are green and B be the event that all 3 marbles are red. So, the required probability is P(A∪B)=P(A)+P(B).
Here, P(A)=C(3,3)C(12,3)=1220
and P(B)=C(5,3)C(12,3)=10220
Hence, P(A∪B)=P(A)+P(B)=11220=120
Answered by
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Answer: 0.063
Step-by-step explanation:
number of all marbles 7+5+4=16. Red on first draw 7/16. Red on second draw 6/15. Red on third draw 5/14. 7/16*6/15*5/14=210/3360 or 0.0625 round up to nearest 1000th=0.063
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