1.

ar contains blue and green marbles. The number of greetnmarbles is 5 more than twicelue. If the probabilitmarbles are there in a jat.green marbleof drawing a blue one at random is 2/7 how many blue and green

Answer»

Let number of blue marbles = nThen number of green marbles= 2n + 5

Probability of drawing blue marble = n/(3n + 5)

As per given conditionn/(3n + 5) = 2/77n = 2(3n + 5)7n - 6n = 10n = 10

Therefore, Number of blue marbles = 10Number of green marbles = 2n + 5 = 2*10 + 5 = 25

let number of blue marbles = x

So, number of green marbles = 2x+5

Therefore, Total number of outcomes = Total number of marbles = n(green)+n(blue) = (2x+5)+(x) = 3x+5

Now, Probability of drawing a blue marble = x / 3x+5......a

given that Probability of drawing a blue marble = 2 / 7

Substituting the value in eq a, we get

(2 / 7) = (x / 3x+5)cross multiplying, 2(3x+5) = 7(x)6x + 10 = 7x7x - 6x = 10x = 10

So, number of blue marbles = x = 10 number of green marbles = 2x+5 =2(10)+5 = 25 total number of marbles = 3x+5 = 3(10)+5 = 35



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