Case Probability Questions

  • Thread starter daewoo
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  • #1
daewoo
25
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Homework Statement


Hey, i need some help with these two probability questions

1)Jeanie is a bit forgetful and if she doesn't make a "to do" list the probability that she forgets some thing is supposed to do is .1. Tomorrow she intends to run three errands and she fails to write them on her hand.
c)what is the probability jeanie remembers at least one of the three errands?

2)In a small city approx 15% of those eligible are called for jury duty in anyone calendar year. People are selected at random from those eligible. The same individual cannot be called more than once in the same year. What is the probability that an eligible person in this city is selected 2 years? 3 years?

Homework Equations


not sure if the second one is a binomial probability distribution


The Attempt at a Solution


For the first question i did:
P(Jeanie remembers first errand but not second or third)
P(jeanie remembers first) X P(jeanie forgets second) x P(jeanie forgets third)
(1-0.1)(0.1)(0.1)= 0.009 I got that, but I'm not too sure if i did it right just hoping someone could double check.

For the second question i really actually don't know how to set this up, so i haven't had much luck.
 

Answers and Replies

  • #2
mattmns
1,118
6
What you did for the first one is wrong.

You have two different ways you can solve the problem.
(i) (Easy way) You could find the probability that she remembers none of the errands, and then the total probability will be 1 minus that number. (The negation of "at least 1" is "less than 1" (in that case it is exactly 0 since you cannot have a negative number of errands).
(ii) (Longer way) Break up into cases. Case 1: She remembers 1 errand (here you have three cases: 1. she remembers first, forgets other two; 2. she remembers second, forgets other two; 3. she remembers third, forgets other two. Then Case 2: She remembers 2 errands (here again you have three cases), and finally Case 3: She remembers all 3 errands. Then you will need to add up all the probabilities, and that would be your answer.

Both ways you should get the answer to be .999


For the second problem. Could you state the exact wording of the problem? Are you wanting the probability that someone will be selected ONCE in two years, or are you looking at the probability of begin selected TWICE in two years?
 
  • #3
daewoo
25
0
oh thanks for the help for the first one,

The second question asks what's the probability that a person will be selected once two years in a row, so he's selected once in the first year, and the second year he is selected again.
 
  • #4
mattmns
1,118
6
What is the probability that he is selected for the first year? What is the probability that he is selected for the second year (does this depend on whether or not he was selected the first year?)?

The answer you get should be: .0225
 

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