Question

A bag contain 5 red marbles,4 green marbles and 1 blue marble. A marble is chosen at random from the bag and not replaced; then a second marble is chose. What is the probability both marbles are green

Answer 1

Answer:

$\frac{2}{15}$

Step-by-step explanation:

GIVEN: A bag contain $5$ red marbles,$4$ green marbles and $1$ blue marble. A marble is chosen at random from the bag and not replaced, then a second marble is chose.

TO FIND: What is the probability both marbles are green.

SOLUTION:

Total marbles in bag $=10$

total number of green marbles $=4$

Probability that first marble will be green $P(A)=\frac{\text{total green marbles}}{\text{total marbles}}$

                                                                    $=\frac{4}{10}=\frac{2}{5}$

Probability that second marble will be green $P(B)=\frac{\text{total green marble in bag}}{\text{total marble in bag}}$

                                                                                    $\frac{3}{9}=\frac{1}{3}$

As both events are disjoint

probability both marbles are green $=P(A)\times P(B)$

                                                           $=\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}$

Hence  the probability both marbles are green is $\frac{2}{15}$