Question

A bag contains 6 blue marbles, 10 red marbles, and 9 green marbles. If two marbles are drawn at random without replacement, what is the probability that two green marbles are drawn? (IT IS NOT 9/25!)

Answer 1

The probability of drawing two green marbles without replacement is
`(3)/(25)`

Explanation:

We are given the following information in the question:

Number of blue marbles = 6

Number of red marbles = 10

Number of green marbles = 9

Total number of marbles = 25

Formula:

`\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}`

`\text{Probability two green marbles are drawn} =\text{Probability of drawing green marble in } 1^(st) \text{ draw}* \text{Probability of drawing green marble in } 2^(nd) \text{ draw}`

`\text{Probability of drawing green marble in } 1^(st) \text{ draw} = \displaystyle(9)/(25)\\\\\text{Probability of drawing green marble in } 2^(nd) \text{ draw} = \displaystyle(8)/(24)`

`\text{Probability two green marbles are drawn} = \displaystyle(9)/(25)* \displaystyle(8)/(24) = (3)/(25)`

Hence, the probability of drawing two green marbles without replacement is
`(3)/(25)`

Answer 2

Total marbles = 6 + 10 + 9 = 25

First pick being green would be 9/25

After picking 1 green, there would be 8 green left and 24 marbles left, so the 2nd pick would be 8/24

The probability of picking both green would be 9/25 x 8/24 = 3/25