1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gorilla in Zoo problem

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Gorilla in a ZOO did not eat whole day, and he is sad. In one moment three kids simultaniusly throw him banana. Probability that first kid is going to throw banana to gorilla is 0,7, second 0,6 and third 0,8.
    If one banana gets to the gorilla he will stodge with probability of 0,4, if two bannanas gets to him with 0,7 and if three, 0,9.

    Find probabilty that gorila will stodge and be happy

    3. The attempt at a solution
    Well, i should try to solve this by making tree diagram. But i dont understand next.
    First i make first branch,

    But the formulation of problem is what causing me problems.
    What does one banana means. Does it means first kids banana, or one of all of those?

    And after i make tree diagram, and suppose it means one bannan = first kids bananna, should i just multiply all probabilities with coresponding and then sum them?
     
  2. jcsd
  3. May 6, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In the 'if one banana reaches the gorilla' onwards, it doesn't matter which bananas. It's just the number of bananas at that point.
     
  4. May 6, 2015 #3
    How can i make tree diagram if i cant distinguish who's bannana is it?
     
  5. May 6, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Why is it essential to make a tree diagram?
    What are the possibilities for the numbers of bananas thrown? What is the probability of each number?
     
  6. May 6, 2015 #5
    Well, i also think of using Bayes formulae. I think of tree diagram because one collegue said he solve it with tree diagram :/
     
  7. May 6, 2015 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For writing simplicity I will call the banana the first kid throws to him, that gets to him with probability 0.7, "banana A", with 0.6, "banana B", and wity probability 0.8, "banana C".

    The probability that exactly one banana gets to him is the sum of the probabilities that exactly one of those bananas get to him and the others don't.
    So: the probability that banana A gets to him but bananas B and C do not is (0.7)(1- 0.6)(1- 0.8)= 0.5(0.4)(0.2)= 0.4. The probability that banana B gets to him but bananas A and C do not is (1- 0.7)(0.6)(1- 0.8)= (0.3)(0.6)(0.2)= 0.36. The probability that banana C gets to him but not bananas A or C do not is (1- 0.7)(1- 0.6)(0.8)= (0.3)(0.4)(0.8)= 0.096. The probability that exactly one banana gets to him is the sum of those: 0.4+ 0.36+ 0.096= 0.856.

    Now, do the same for "exactly two bananas reach him" and "all three bananas reach him", multiply each of those by the probability he will "stodge" and add.

    (And I can't help but wonder what "stodge" means!)
     
  8. May 6, 2015 #7
    So, what i have done is next.
    A={ Gorilla has studged}
    Hi ={ i kids hit gorilla}

    P(H0)= ... doesnt matter because later in formuale its multiplied with 0.

    P(H1)=7/10 * 4/10 * 2/10 *(3choose1)
    P(H2)=7/10 * 6/10 * 2/10 * (3choose2)
    P(H3)= 7/10 * 6/10 * 8/10

    P(A|H1)=0,4, P(A|H2)=0,7, P(A|H3)=0,8
    P(A) = SUM(P(Hi)*P(A|Hi))= 0,5124
     
  9. May 6, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's wrong. You've taken the probability that a particular banana is thrown, and not the other two, and multiplied by 3. But each has a different probability.
     
  10. May 6, 2015 #9
    Mhm, how do i include possibilities of all being thrown?
     
  11. May 6, 2015 #10
    Hey hallsofly! I thank you for replaying and for taking your time to write me that. I appreciated.
    Stogde mean be full, full-fed.

    My question now is next. I understand what you have done, i i will do now for options where two and three bananas get to him, but problem i last sentence, where i am asked to count probability where is he full-fed and happy. How do i do that? I really dont understand what it means.

    Does it mean i need to count for two and three bananas and the one with highest percentage of getting to him is going to make him happy and full?

    Thank you again for quick responses. You are the best!
     
  12. May 6, 2015 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You said "If one banana gets to the gorilla he will stodge with probability of 0,4, if two bannanas gets to him with 0,7 and if three, 0,9."

    If "stodge" means "become happy", then multiply 0.4 times the probability exactly one banana is thrown to him, 0.7 times the probability exactly two bananas are thrown to him, 0.9 times the probability three are thrown to him, and add those.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Gorilla in Zoo problem
  1. Summation problem (Replies: 2)

  2. Vector problem (Replies: 15)

Loading...