Doubt from Probability

#1

Aman, six of his friends, and a monkey are standing at different vertices of a big cube. Aman has a banana and is a neighbour to the monkey. On each turn, whoever has the banana throws it at random to one of his three neighbours. If the monkey gets the banana he eats it. What is the probability that Aman is the one who feeds the monkey?

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#2

3/7

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#3

Agreed Sir!
We get a GP with common ratio as \frac{2}{9} and first term \frac{1}{3}
Basically we need to first see how this works by taking a few cases... We then see that the banana will reach Aman in even no. of steps and for each added step we have two more possibilities... Indeed a nice problem involving logic

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#4

Answer is 7\12 @Sourav_Dey @Mayank_2019_1 sir

#5

Okay, I understood my mistake, I did not consider the fact that if monkey gets the banana then the process will stop... Basically 2 of his friends will have only two other places to pass the banana if Aman has to give the banana to the monkey.
Now, that 1/9 term is okay but for the multiplication by 2 there will be more cases
Wait will have to see the question

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#6

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#7


The Cube is shown above, where A is aman and M is the monkey. Let p be the probability that Aman feeds the monkey given that he is holding the banana. For each non-monkey neighbor of aman, let q be the probability that aman feeds the monkey
given that the neighbor is holding the banana. (By symmetry, it is the same probability for each neighbor.) For each non-aman neighbor of the monkey, let r be the probability that aman feeds the monkey given that the neighbor is holding the banana. (Again, by symmetry, it is the same probability for each of the monkey's
other neighbors.) Finally, let s be the probability that aman feeds the monkey given that the person opposite from aman is holding the banana. Note also that if the person who is opposite the monkey has the banana, then the probability that aman feeds the monkey is 1 \over 3 , by symmetry, since each of the monkey's three neighbors is then equally likely to be the feeder. These probabilities are labeled on the diagram above. When aman has the banana, he has probability 1 \over 3 of immediately feeding the monkey, and probability 2 \over 3 of passing it to a neighbor. Therefore,
p = 1\over 3 + 2 \over 3 q
When one of aman's non-monkey neighbors has the banana, she has probabilty 1\over 3 of throwing
the banana back to aman, probability 1 \over 3 of throwing the banana to one of the monkey's other neighbors, and probability 1 \over 3 of throwing the banana to the person opposite the monkey. Therefore,
q= 1 \over 3 p + 1 \over 3 r + 1 \over 9
When one the monkey's other neighbors has the banana, he has probability 1 \over 3 of feeding the monkey (in which case aman does not feed the monkey), probability 1 \over 3 of throwing the banana to one of aman’s neighbors, and probability 1 \over 3 of throwing the banana to the person
opposite aman. Therefore,
r = 1 \over 3 q + 1 \over 3 s.
Finally, when the person opposite aman has the banana, she has probability 2 \over 3
of throwing it to one of the monkey’s non-aman neighbors, and probability 1 \over 3
of throwing it to the person opposite to the monkey. Therefore,
s = 2 \over 3 r + 1 \over 9
Thus, we have the system of equation
p= 1 \over 3 + 2 \over 3 q
q= 1\over 3 p + 1 \over 3 r + 1 \over 9
r= 1 \over 3 q + 1 \over 3 s
s= 2 \over 3 r + 1 \over 9
Solving this system of equation gives p= 7 \over 12 and q= 3 \over 8 . So the probability that aman feeds the monkey is 7 \over 12

Thanks.

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