Probability of Gorilla Stodging with Throwing Bananas

[diracdelta]

Homework Statement

[/B]
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 probability that gorila will stodge and be happy

The Attempt at a Solution

Well, i should try to solve this by making tree diagram. But i don't 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?

 

[haruspex]

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.

 

[diracdelta]

How can i make tree diagram if i can't distinguish who's bannana is it?

 

[haruspex]

How can i make tree diagram if i can't distinguish who's bannana is it?

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?

 

[diracdelta]

Well, i also think of using Bayes formulae. I think of tree diagram because one colleague said he solve it with tree diagram :/

 

[HallsofIvy]

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!)

 

[diracdelta]

So, what i have done is next.
A={ Gorilla has studged}
H_i ={ i kids hit gorilla}

P(H0)= ... doesn't 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(H_i)*P(A|H_i))= 0,5124

 

[haruspex]

P(H1)=7/10 * 4/10 * 2/10 *(3choose1)

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.

 

[diracdelta]

Mhm, how do i include possibilities of all being thrown?

 

[diracdelta]

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!)

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 don't 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!

 

[HallsofIvy]

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.