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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.

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.

Answer

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piratepower12
7 months ago

Светило науки - 38 ответа - 0 помощи
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%

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.