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 0.05By the third day of a particular week, 2 accidents have already occurred in the intersection. What is the probability that there will be less than a total of 4 accidents during that week? 1.00 0.90 0.85 0.70 0.50

The problem is asking for the probability that there will be less than 4 total accidents in the week, given that there have already been 2 accidents in the first 3 days.

This means we are looking for the probability that there will be either 0 or 1 more accidents in the remaining 4 days of the week.

The probability distribution for the number of accidents in a week is given as:

x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05

We can ignore the probabilities for x = 2, 3, 4, and 5, because these represent more accidents than we are interested in.

So, we are interested in the probabilities for x = 0 and x = 1.

P(X = 0) = 0.20 P(X = 1) = 0.30

The probability that there will be less than a total of 4 accidents during the week is the sum of these probabilities, which is 0.20 + 0.30 = 0.50.

So, the answer is 0.50.

To solve this problem, we need to find the probability that no more accidents occur in the remaining 4 days of the week, given that 2 accidents have already occurred.

The probability distribution of X is given as:

x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05

We are interested in the probability that X is less than 4, given that 2 accidents have already occurred. This is equivalent to finding the probability that X is 2 or 3.

P(X < 4 | 2 accidents have already occurred) = P(X = 2) + P(X = 3)

From the given probability distribution, we know that P(X = 2) = 0.20 and P(X = 3) = 0.15.

So, P(X < 4 | 2 accidents have already occurred) = 0.20 + 0.15 = 0.35

Therefore, the probability that there will be less than a total of 4 accidents during that week, given that 2 accidents have already occurred, is 0.35 or 35%. This is not one of the options given, so there may be a mistake in the problem or in the provided options.