Question

A bag contains 1 blue, 2 green, and 3 red marbles, as shown. What is the probability of drawing a green marble out of the bag without looking?

Answer 1

ANSWER

$P(G)= \frac{1}{3}$

EXPLANATION

The number of green marbles in the bag is

$n(G) = 2$

The total number of marbles in the bag is

$n(S)=1+2+3 = 6$

The probability of selecting a green marble from the bag without looking is

$P(G)= \frac{n(G)}{n(S)}$

Substitute the values to get,

$P(G)= \frac{2}{6}$

$P(G)= \frac{1}{3}$

Answer 2

Answer:

The probability of drawing a green marble out of the bag without looking = 1/3

Step-by-step explanation:

It is given that, a bag contains 1 blue, 2 green, and 3 red marbles

Therefore total number of marble in the  bag = 1 + 2 + 3 = 6

To find the probability

Total number of marble = 6

Number of green marble  = 2

The probability of drawing a green marble = 2/6 = 1/3