Question

Two marbles are drawn without replacement from a box with 3 white, 2 green, 2 red, and 1 blue marble. Find the
probability.
Both marbles are green.

Answer 1

Answer:

The probability that both marbles are green is $\bf\displaystyle\frac{1}{28}$.

Step-by-step explanation:

To find the probability of drawing 2 green marbles, we create 2 events where:

1. A = 1st marble is green
2. B = 2nd marble is green

Hence, the event of drawing 2 green marbles on both 1st and 2nd is P(A and B) or P(A∩B).

Since the outcome of event B depends on event A, then these events are dependent events (conditional probability), where:

$\boxed{P(A\cap B)=P(A)\times P(B|A)}$

Event A:

• Total number of marbles $(n(S)_A)$ = 3 + 2 + 2 + 1 = 8
• Total number of green marble $(n(A))$ = 2

$\displaystyle P(A)=\frac{n(A)}{n(S)_A}$

         $\displaystyle =\frac{2}{8}$

Event B:

• Total number of marbles $(n(S)_B)$ = 8 - 1 = 7 (1 marble was taken at event A)
• Total number of green marbles $(n(B))$ = 2 - 1 = 1 (1 green marble was taken at event A)

$\displaystyle P(B|A)=\frac{n(B)}{n(S)_B}$

            $\displaystyle =\frac{1}{7}$

Therefore:

$P(A\cap B)=P(A)\times P(B|A)$

               $\displaystyle=\frac{2}{8} \times\frac{1}{7}$

               $\bf\displaystyle=\frac{1}{28}$