Question

The probability that John goes to the show is 1/3. If he goes to the show, the probability that he sees a python is 2/5. If he does not go to the show the probability that he sees a python is 1/8. Find tha probability that;
i. John goes to the show but she doesn’t see a python
ii. John sees a python elsewhere

Answer 1

Answer on Question #56399 – Math – Statistics and Probability

The probability that John goes to the show is 1/3. If he goes to the show, the probability that he sees a python is 2/5. If he does not go to the show the probability that he sees a python is 1/8. Find the probability that;

i. John goes to the show but she doesn't see a python

ii. John sees a python elsewhere

Solution

Let $A$ be the event "John goes to the show" and $B$ be the event "John sees a python". Then

$$P(A) = \frac{1}{3}; P(B|A) = \frac{2}{5}; P(B|\bar{A}) = \frac{1}{8}.$$

i. The probability that John goes to the show but she doesn't see a python is

$$P(A \text{ and } \bar{B}) = P(A)P(\bar{B}|A) = P(A)\left(1 - P(B|A)\right) = \frac{1}{3}\left(1 - \frac{2}{5}\right) = \frac{1}{3} \cdot \frac{3}{5} = \frac{1}{5}.$$

ii. The probability that John sees a python elsewhere is

$$P(\bar{A} \text{ and } B) = P(\bar{A})P(B|\bar{A}) = \left(1 - P(A)\right)P(B|\bar{A}) = \left(1 - \frac{1}{3}\right)\frac{1}{8} = \frac{2}{3} \cdot \frac{1}{8} = \frac{1}{12}.$$

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