Question
Download Solution PDFIn an experiment, it is found that the probability that a released bird will come back is \(\rm \dfrac {2}{5}\). It is also found that when a bird comes back, the probability that the bird will stay is \(\rm \dfrac {1}{3}\). In this experiment, what is the probability that a released bird will come back and fly away?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
- P(not E) = 1 - P(E).
- Probability of a Compound Event [(A and B) or (B and C)] is calculated as:
P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)] ('and' means '×' and 'or' means '+')
Calculation:
It is given that P(Bird comes back) = \(\rm \dfrac {2}{5}\) and P(Bird stays) = \(\rm \dfrac {1}{3}\).
The probability that the bird does not stay (flies away) after coming back = \(\rm 1-\dfrac {1}{3}=\dfrac 2 3\).
The probability of the event in question is:
P(Bird comes back) AND P(Bird does not stay)
= P(Bird comes back) × P(Bird does not stay)
= \(\rm \dfrac 2 5 \times \dfrac {2}{3}=\dfrac {4}{15}\).
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