Science:Math Exam Resources/Courses/MATH102/December 2012/Question B 06
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Question B 06
There are six neurons connected to a muscle. If four or more of the neurons fire at the same time, the muscle contracts. In any millisecond, each neuron has a 0.3 probability of firing. What is the probability that the muscle contracts in any particular millisecond? You don't have to simplify your answer.
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Compute the probability that exactly 6 neurons fire, then exactly 5 then exactly 4 and add these up. Don't forget that there are many combinations of exactly 5 neurons firing and exactly 4 neurons firing.
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(In what follows, the notation P(outcome) should be read as "the probability that 'outcome' occurs.")
In order to solve this problem, we need to break it down into pieces. We are asked to find:
A muscle contracts if 4 or more neurons fire. Since there are 6 neurons total we can rewrite this probability as:
When you have different outcomes joined by an "or", you can separate the outcomes into individual probabilities and add them together. So
Hopefully the problem is more manageable now - we're just computing three individual probabilities and then add them up. Let's start with the last one, because it's the easiest. (For ease, we will be calling the six neurons a, b, c, d, e and f.)
Recall that when probabilities are joined by an "and", you multiply them. Since the probability of each neuron firing is .3, this means that
Now onto the next probability, P(5 neurons fire). Suppose I had a list of 5 specific neurons I wanted to fire, say a and b and c and d and e. Since only 5 are firing, this also means the sixth neuron f does not fire, which has a probability of .7. The probability of this specific combination occurring involves multiplying the probability for each neuron firing/not firing, which gives (.3)5(.7). However, this is not the only combination of 5 neurons that can fire - we need to include all possible combinations of 5 neurons. This can be done using the binomial . Since the same (.3)5(.7) probability will occur for each list of 5 neurons, our overall probability is
Finally, computing P(4 neurons fire), we use the same method as the previous probability. The probability that a specific group of 4 neurons will fire is (.3)4, the probability that the other two neurons do not fire is (.7)2, and the number of groups of 4 neurons is found by the binomial , giving
Since we don't need to simplify any further, we can simply add our probabilities together as shown above to get a final answer of: