Question

given, ( p(e) = 0.105 ), ( p(ecup f) = 0.351 ), ( p(ecap f) = 0.098 ), what is ( p(f) =? ) note: enter your answer in the box below without rounding it. question 7 the chicago bears football team has a probability of 0.63 to win the next game, and a probability of 0.11 to tie the next game. what is the probability for the team to lose the next game? note: please enter your final answer in full without rounding it.

Explanation:

Step1: Use probability formula

The sum of probabilities of all mutually - exclusive events (win, tie, lose) is \(1\). Let \(P(\text{win}) = 0.63\), \(P(\text{tie})=0.11\), and \(P(\text{lose})\) be the probability of losing. Then \(P(\text{win})+P(\text{tie})+P(\text{lose}) = 1\).

Step2: Solve for \(P(\text{lose})\)

Rearrange the formula: \(P(\text{lose})=1-(P(\text{win}) + P(\text{tie}))\). Substitute the values: \(P(\text{lose})=1-(0.63 + 0.11)\). First, calculate \(0.63+0.11 = 0.74\). Then \(P(\text{lose})=1 - 0.74\).

Answer:

\(0.26\)