The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:x 0 1 2 3 4 5P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05What is the probability that in a given week there will be at most 3 accidents? 0.70 0.85 0.35 0.15 1.00
Question
The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:x 0 1 2 3 4 5P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05What is the probability that in a given week there will be at most 3 accidents? 0.70 0.85 0.35 0.15 1.00
Solution
The probability that there will be at most 3 accidents in a given week is found by adding the probabilities of having 0, 1, 2, or 3 accidents.
From the given probability distribution:
P(X = 0) = 0.20 P(X = 1) = 0.30 P(X = 2) = 0.20 P(X = 3) = 0.15
So, P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.20 + 0.30 + 0.20 + 0.15 = 0.85
Therefore, the probability that there will be at most 3 accidents in a given week is 0.85.
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