Respuesta :
Answer:
a) 0.3 = 30% probability of drawing a white chip.
b) 0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
a. P(W), the probability of drawing a white chip.
1/3(one white out of 3) of 0.3(from B1).
1/2(two white out of 4) of 0.2(from B2).
1/5(one white out of 5) of 0.5(from B3). So
[tex]P(W) = 0.3333*0.3 + 0.5*0.2 + 0.2*0.5 = 0.3[/tex]
0.3 = 30% probability of drawing a white chip.
b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.
The probability of drawing a white chip from B1 is 1/3 out of 0.3, so:
[tex]P(B1 \cap W) = 0.3\frac{1}{3} = 0.1[/tex]
Then the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.3} = 0.3333[/tex]
0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.
