meredith48034
Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

NO LINKS!!

Mr. Magoo is forgetful and frequently leaves the lights on his car. There is a 40% probability he will leave his on any randomly selected day. There are 5 days in a normal school week. NO MULTIPLE CHOICE!

a. Find the probability he leaves his lights on all 5 days this week.

b. Find the probability he remembers to turn his lights off all 5 days this week.

c. Find the probability Mr. Magoo leaves his lights on at least one day this week. ​

Sagot :

a. (40%^5)=0.4^5=0.01024, which is 1.024%

b. the probability that he turns is light off is the probability that he doesn't turn his light on, which is 1-0.4=0.6.

0.6^5=0.07776, or 7.776%

c. 1-0.07776=0.92224, or 92.224%

luclaporte2

Answer:

a. 40% = 0.4 → (0.4)⁵ = 0.01024.. ≈ 1.02%.

b. (1 - 0.4) → (0.6)⁵ = 0.07776.. ≈ 7.78%

c. (0.4)¹ → 0.4 = 40%

Explanation:

decimal × 100% = percent.

Since these independent events either occur or do not occur, we can use binomial distribution to determine the theoretical probability of either outcome with a given number of successes or failures (total - successes) or by probability → 1 - success probability = failure probability.

Since the problem is either looking for all or none, we can narrow n! / (n - x)!x! p^x q^n-x down to (probability)^(days). All the rest cancels out.

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.