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Clarification for Probability problem

  1. Oct 10, 2004 #1
    Please forgive my statistical ignorance! I have a couple of problems in my Stats class that I would like anyone to review and let me know if I am on the right track...

    You have applied for a job at Store A and also Store B. You estimate the probabilities of getting the job at the stores A and B are 0.1 and 0.7 respectively. You also estimate that the probability of getting both jobs is only 0.04.

    What is the probability that you get at least one of the jobs?

    a) 0.04
    b) 0.72
    c) 0.8
    d) 0.76

    I chose c) 0.8 (by adding the probabilities of A and B together)

    What is the probability that you will get exactly one of the jobs?

    a) 0.72
    b) 0.04
    c) 0.76
    d) 0.8

    I chose c) 0.76 (by adding the probabilities of A and B together, then subtracting the probability of getting both jobs)

    Could someone/anyone look at this and tell me if I am on the right track, please? I am having a terrible time trying to understand this theory! Thanks in advance for any help you can offer me.
  2. jcsd
  3. Oct 10, 2004 #2


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    No, that "0.7" probability of getting job B includes the probability of get job A also. P(A or B)= P(A)+ P(B)- P(A and B).

    Probability of getting job A but NOT job B is 0.1- 0.04= 0.06.
    Probability of getting job B but NOT job A is 0.7- 0.04= 0.66.

    The probability of gett exactly one job is the same as the probability of getting A but not B OR getting B but not A- since those can't both happen, the probability is 0.06+ 0.66= 0.72.

  4. Oct 10, 2004 #3
    EDIT: Looks like my post below is incorrect. I'm leaving it so I can figure out where I went wrong as well.

    QUESTION EDIT: So the .1 and .7 are the probabilities of getting a job at each store if you go to both stores not individually? So basically, if you go to store A & B you have a .1 chance at A, .7 at B, and a persumed .2 failure rate. Your chance at getting both jobs (if you go to both stores) is .4 when you add the chances and divide by 2?

    The percentages are different because you are going to both stores? Applied to a realistic situation perhaps applying at both gives you an advantage because the manager owns both stores? He wants a worker at both places to minimize micro-managing and because the stores are similiar allowing for less training requirements?

    Also, if the chance at getting a job at A alone was .1 and B .7, then the figuring below would be correct?

    Sorry in advance over my untechnical answers that may be incorrect. I might be able to give you ideas to work with though. After looking over your answers I think they are correct.

    Part 1

    10/100 + 70/100 = 80% or .8 = Your chance of getting a single job.
    Since getting both jobs is dependent on getting a single job, the .04 would not apply.

    If got a single job you would have a job. Afterwards, you have a job and thus have the one job required. After you have one job having two jobs doesn't matter.

    Part 2: That seems right to me, I'm not sure how to explain it without quoting your post.
    Last edited: Oct 10, 2004
  5. Oct 10, 2004 #4
    Thank you both so much for your help and input!! What you said actually makes sense and I am grateful especially for your explanations as to where I was right or went wrong.
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