Probabilityistic
A bag contains blue and green marbles. If 5 green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be 75/83 . If instead 7 blue marbles are added to the bag, the probability of drawing a blue marble would be 3/19 . What was the number of blue marbles in the bag before any changes were made?
The answer is 8.
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1 solution
Suppose there were B blue marbles and G green marbles initially. If 5 green marbles are removed, number of remaining green marbles is G-5 and total number of remaining marbles is B+G-5. In this case, the probability of drawing a green marble is (G-5)/(B+G-5). It is stated that this probability of drawing green marbles is 75/83. Hence, (G-5)/(B+G-5) = 75/83. ----- equation (1)
If 7 blue marbles are added, number of remaining blue marbles is B+7 and total number of remaining marbles is B+G+7. In this case, the probability of drawing a green marble is (B+7)/(B+G+7). It is stated that this probability of drawing blue marbles is 3/19. Hence, (B+7)/(B+G+7) = 3/19. ----- equation (2)
On solving these two equations, we get B=8, which is the answer.