Question

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\) and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F ?\) b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F) ?\) A Venn diagram might help.)

Short answer

The probabilities are: 0.55 for stopping at at least one light, 0.45 for not stopping at either light, 0.4 for stopping at exactly one light, and 0.25 for stopping only at the first light.

Step-by-step solution

Probability of Stopping at At Least One Light

To determine the probability that the individual must stop at at least one light, which is denoted as \(E \cup F\), we use the sum rule for probabilities which states that \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). In substitution, we get \(P(E \cup F) = 0.4 + 0.3 - 0.15 = 0.55.\)

Probability of Not Stopping at either Light

To find the probability that the individual doesn't have to stop at either light, we need to find the complement of the event calculated in Step 1. This is given by \(P(\sim(E \cup F)) = 1 - P(E \cup F)\). Substituting the calculated value, we get \(P(\sim(E \cup F)) = 1 - 0.55 = 0.45.\)

Probability of Stopping at Exactly One Light

The probability that the individual must stop at exactly one of the two lights is given by the sum of probability of stopping at light one only and light two only. This is mathematically expressed as \(P(E \sim F) + P(\sim E F) = [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]\). Substituting in the given values, we get \(P(E \sim F) + P(\sim E F) = [0.4 - 0.15] + [0.3 - 0.15] = 0.4\).

Probability of Stopping Just at the First Light

The probability that the individual stops just at the first light is given by the probability of event \(E\) minus the joint probability of both events \(E\) and \(F\) occurring. This can be mathematically presented as \(P(E \cap \sim F) = P(E) - P(E \cap F)\). Substituting the given values, we have \(P(E \cap \sim F) = 0.4 - 0.15 = 0.25\).