Questions: What is the probability of getting a parking ticket on campus on at least one out of 4 days without a parking pass if the chance of not getting a ticket on a particular day when a parking pass isn't present is 0.15? The probability is approximately %. (Simplify your answer. Round to one decimal place as needed.)

Solution Steps

We need to find the probability of getting a parking ticket on campus on at least one out of 4 days without a parking pass. The probability of not getting a ticket on a particular day when a parking pass isn't present is given as \( P(\text{no ticket}) = 0.15 \). Therefore, the probability of getting a ticket on a particular day is:

\[ P(\text{ticket}) = 1 - P(\text{no ticket}) = 1 - 0.15 = 0.85 \]

To find the probability of getting at least one ticket over 4 days, we first calculate the probability of getting no tickets at all over these 4 days. This can be expressed using the binomial distribution formula for \( x = 0 \):

\[ P(X = 0) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 4 \) (number of days),
• \( x = 0 \) (number of tickets),
• \( p = 0.85 \) (probability of getting a ticket),
• \( q = 0.15 \) (probability of not getting a ticket).

Calculating this gives:

\[ P(X = 0) = \binom{4}{0} \cdot (0.85)^0 \cdot (0.15)^4 = 1 \cdot 1 \cdot (0.15)^4 = 0.00050625 \]

Now, we can find the probability of getting at least one ticket over the 4 days:

\[ P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00050625 \approx 0.99949375 \]

To express this probability as a percentage, we multiply by 100:

\[ P(X \geq 1) \times 100 \approx 99.949375 \% \]

Rounding this to one decimal place gives:

\[ P(X \geq 1) \approx 100.0 \% \]

Final Answer