Question

A neuron N is connected with three other neurons N1 , N2, and N3 capable of producing a stimulus by firing, {N1 , N2, N3} −→ N.

The probabilities of firing for neurons N1, N2, and N3 are 0.2 each and the actions of these three neurons are simultaneous and mutually independent.

Neuron N receives a stimulus if at least one of N1 , N2, and N3 fires.

Neuron N will fire with probability 0.9 if a stimulus is present and with probability 0.05 if a stimulus is not present.

Using Bayes Theorem, what is the probability that N fires if N1 did not fire?

Answer 1

a) Neuron receives a stimulus = P(at least one of N1 , N2, and N3 fires)

1- P(None of N1 , N2, and N3 fires) = 1- ( 0.8 * 0.8 *0.8) = 0.488

b) P(N fire | stimulus ) = 0.9

P(N fire | no stimulus ) = 0.05

P(N fire) = P(N fire | stimulus ) * P(stimulus) + P(N fire | no stimulus )* P(no stimulus)

= 0.9 * 0.488 + 0.05 * 0.512 = 0.4648

c)

Probability of stimulus =[(0.2*0.8*0.8) + (0.8*0.8*0.2) +(0.2*0.8*0.2)] = 0.288

P(N fire) = P(N fire | stimulus ) * P(stimulus) + P(N fire | no stimulus )* P(no stimulus)

= 0.9 * 0.288 + 0.05 * 0.712 = 0.2948

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