Question

Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking two marbles from the jar without replacing the first one. What is the probability of getting a green marble and a red marble? (Enter your probability as a fraction. Hint: How is this exercise different from finding the probability of getting a green marble first and a red marble second?)

Answer 1

Answer:

$\frac{5}{13}$

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

$\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

Since, here replacement is not allowed,

Thus, the probability of getting a green marble and a red marble

= first red and second green + first green second red

$=\frac{3}{13}\times \frac{10}{12}+\frac{10}{13}\times \frac{3}{12}$

$=2\times \frac{3}{13}\times \frac{10}{12}$

$=\frac{30}{78}$

$=\frac{5}{13}$

Note : The probability of getting a green marble first and a red marble second

= $\frac{10}{13}\times \frac{3}{12}$

$=\frac{30}{156}$

$=\frac{5}{26}$