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 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 What 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
Accepted Solution
A:
Answer: 0.70Step-by-step explanation:Given : 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 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05Using the above probability distribution , the the probability that in a given week there will be at most 3 accidents is given by :_[tex]P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70[/tex]Hence, the required probability = 0.70