+ Reply to Thread
Results 1 to 2 of 2

Thread: Help please

  1. #1

    Help please




    An Olympic archer is able to hit a bull's-eye 80%of the time. Assume each shot is independent. If she shoots 6 arrows, what's the probability of her hitting her first bull's-eye on the fourth or fifth arrow?

    My question is this: do I find the probability of getting the first bull's-eye on the fourth shot, then find the probability of getting the first bull's-eye on the fifth shot, the multiply the two probabilities together? If not what do you do?

  2. #2
    Points: 1,097, Level: 17
    Level completed: 97%, Points required for next Level: 3

    Location
    Philadellphia, PA
    Posts
    68
    Thanks
    1
    Thanked 20 Times in 18 Posts

    Re: Help please


    P(miss) = .2
    P(hit) = .8

    Let first bull's-eye on the fourth shot = P(A)
    Let first bull's-eye on the fifth shot = P(B)

    P(A) = .2 * .2 * .2 * .8 = 0.0064
    P(B) = .2 * .2 * .2 * .2 * .8 = 0.00128

    Because A and B are mutually exclusive, P(A and B) = 0

    P(A or B) = P(A) + P(B) - P(A and B) = .0064 + 0.00128 - 0 = 0.00768

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats