Answer: 0.70
Step-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 0.05
Using 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
