Jeanie is a bit forgetful, and if she doesn\'t make a \"to do\" list, the probab

Question

Jeanie is a bit forgetful, and if she doesn't make a "to do" list, the probability that she forgets something she is supposed to do is .8. Tomorrow she intends to run three errands, and she fails to write them on her list.

(a) What is the probability that Jeanie forgets all three errands?


What assumptions did you make to calculate this probability?
Events are  ---Select--- dependent independent .

(b) What is the probability that Jeanie remembers at least one of the three errands?


(c) What is the probability that Jeanie remembers the first errand but not the second or third?

Explanation / Answer

P(Forgetting) = 0.8 and P(Remembering) = 1-0.8 = 0.2

The Binomial Probability is given by nCr * pr * q (n-r) , where nCr = n!/[(n-r)!*r!]

(a) P(Forgetting All 3 errands) = 3C3 * (0.8)3(0.2)0 = 0.512

Events here are independant

(b) P(Remembering at least 1) = P(Remembering 1) + P(Remembering 2) + P(Remembering 3)

Remembering 1 = 3C1 * (0.2)1 * (0.8)2 = 3 * 0.2 * 0.64 = 0.384

Remembering 2 = 3C2 * (0.2)2 * (0.8)1 = 3 * 0.04 * 0.8 = 0.096

Remembering 3 = 3C3 * (0.2)3 * (0.8)0 = 1 * 0.008 * 1 = 0.008

Therefore the required probability = 0.3840 + 0.0960 + 0.0008 = 0.488

This is also equal to 1 - P(Remembering 0 errands) = 1 - 0.512(which we calculated in (a)) = 0.488

(c) Here we are specific about errand 1, 2 and 3. So probability of remembering the first = 0.2, and not remembering the 2nd and the 3rd is = 0.8 each. Therefore the required probability = 0.2 * 0.8 * 0.8 = 0.128

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