Engineering
Mathematics
Conditional Probability
Question
A random variable X has the probability distribution :
| X : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) : | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
For the events E = {X is a prime number} and F = {X < 4}, the probability P(E∪F) is
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Solution
P(E∪F) = P(E) + P(F) − P(E∩F))
= (0.23 + 0.12 + 0.2 + 0.07) + (0.15 + 0.23 + 0.12) − (0.23 + 0.12)
= 0.62 + 0.15
= 0.77
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