Respuesta :
Answer:
0.2275 = 22.75% probability that you actually won that round
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.
In this question:
Event A: Fireworks going off
Event B: You won
Probability of fireworks going off.
100% of 1/35 = 0.0286(when you win)
10% of 34/35 = 0.9714(you lost). So
[tex]P(A) = 0.0286 + 0.1*0.9714 = 0.12574[/tex]
Probability of you winning and fireworks going off:
100% of 1/35, so [tex]P(A \cap B) = 0.0286[/tex]
If you failed to see the outcome of a round, but you see the fireworks going off, then what is the probability that you actually won that round?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0286}{0.12574} = 0.2275[/tex]
0.2275 = 22.75% probability that you actually won that round
