meredith48034
Answered

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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. ​

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.

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