Questions: x 0 1 2 3 4 5 P(X) 0.10 0.20 0.45 0.15 0.05 0.05 Using the above information, the probability of less than 3 accidents is A. 0.15 B. 1.00 C. 0.90 D. 0.75

Solution Steps

We are given a discrete probability distribution for the random variable \( X \) representing the number of accidents. The values and their corresponding probabilities are as follows:

\[ \begin{array}{c|c} x & P(X) \\ \hline 0 & 0.10 \\ 1 & 0.20 \\ 2 & 0.45 \\ 3 & 0.15 \\ 4 & 0.05 \\ 5 & 0.05 \\ \end{array} \]

To find the probability of less than 3 accidents, we need to sum the probabilities for \( X = 0, 1, \) and \( 2 \):

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

Substituting the values:

\[ P(X < 3) = 0.10 + 0.20 + 0.45 = 0.75 \]

We compare the calculated probability \( P(X < 3) = 0.75 \) with the provided options:

• A. \( 0.15 \)
• B. \( 1.00 \)
• C. \( 0.90 \)
• D. \( 0.75 \)

The correct option that matches our calculation is \( D \).

Final Answer