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 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

Answer 1

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 :_

$P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70$

Hence, the required probability = 0.70