Question

A bag contains 3 red marbles, 4 blue marbles and 8 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that all three marbles drawn will be blue?​

Answer 1

Answer:

The probability that all three marbles drawn will be blue = 0.9%.

Step-by-step explanation:

To find the probability of drawing 3 blue marbles, we create 3 events where:

1. A = 1st marble is blue
2. B = 2nd marble is blue
3. C = 3rd marble is blue

Hence, the event of drawing 3 blue marbles on all 1st, 2nd & 3rd event is P(A and B and C) or P(A∩B∩C).

Since the outcomes of event B and C depend on their previous events, then these events are dependent events (conditional probability), where:

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

Event A:

• Total number of marbles $(n(S)_A)$ = 3 + 4 + 8 = 15
• Total number of blue marble $(n(A))$ = 4

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

        $\displaystyle =\frac{4}{15}$

Event B:

• Total number of marbles $(n(S)_B)$ = 15 - 1 = 14 (1 marble was taken at event A)
• Total number of blue marble $(n(B))$ = 4 - 1 = 3 (1 blue marble was taken at event A)

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

        $\displaystyle =\frac{3}{14}$

Event C:

• Total number of marbles $(n(S)_C)$ = 15 - 1 - 1 = 13 (2 marbles was taken at event A & B)
• Total number of blue marble $(n(C))$ = 4 - 1 - 1 = 2 (2 blue marbles was taken at event A & B)
• $\displaystyle P(C|B|A)=\frac{n(C)}{n(S)_C}$

        $\displaystyle =\frac{2}{13}$

Therefore:

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

                      $\displaystyle=\frac{4}{15} \times\frac{3}{14} \times\frac{2}{13}$

                      $\displaystyle=\frac{4}{455}$

                      $=0.9\%$