Respuesta :
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:
- A = 1st marble is blue
- B = 2nd marble is blue
- 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:
[tex]\boxed{P(A\cap B\cap C)=P(A)\times P(B|A)\times P(C|B|A)}[/tex]
Event A:
- Total number of marbles [tex](n(S)_A)[/tex] = 3 + 4 + 8 = 15
- Total number of blue marble [tex](n(A))[/tex] = 4
[tex]\displaystyle P(A)=\frac{n(A)}{n(S)_A}[/tex]
[tex]\displaystyle =\frac{4}{15}[/tex]
Event B:
- Total number of marbles [tex](n(S)_B)[/tex] = 15 - 1 = 14 (1 marble was taken at event A)
- Total number of blue marble [tex](n(B))[/tex] = 4 - 1 = 3 (1 blue marble was taken at event A)
[tex]\displaystyle P(B|A)=\frac{n(B)}{n(S)_B}[/tex]
[tex]\displaystyle =\frac{3}{14}[/tex]
Event C:
- Total number of marbles [tex](n(S)_C)[/tex] = 15 - 1 - 1 = 13 (2 marbles was taken at event A & B)
- Total number of blue marble [tex](n(C))[/tex] = 4 - 1 - 1 = 2 (2 blue marbles was taken at event A & B)
- [tex]\displaystyle P(C|B|A)=\frac{n(C)}{n(S)_C}[/tex]
[tex]\displaystyle =\frac{2}{13}[/tex]
Therefore:
[tex]P(A\cap B\cap C)=P(A)\times P(B|A)\times P(C|B|A)[/tex]
[tex]\displaystyle=\frac{4}{15} \times\frac{3}{14} \times\frac{2}{13}[/tex]
[tex]\displaystyle=\frac{4}{455}[/tex]
[tex]=0.9\%[/tex]
