Science:Math Exam Resources/Courses/MATH102/December 2012/Question B 06

MATH102 December 2012

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
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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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (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: P(muscle contracts) A muscle contracts if 4 or more neurons fire. Since there are 6 neurons total we can rewrite this probability as: P(muscle contracts) = P(4, 5, or 6 neurons fire) When you have different outcomes joined by an "or", you can separate the outcomes into individual probabilities and add them together. So P(4, 5, or 6 neurons fire) = P(4 neurons fire) + P(5 neurons fire) + P(6 neurons fire) 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.) P(6 neurons fire) = P(a and b and c and d and e and f fire) Recall that when probabilities are joined by an "and", you multiply them. Since the probability of each neuron firing is .3, this means that P(6 neurons fire) = (.3)6 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 ${\binom {6}{5}}$. Since the same (.3)5(.7) probability will occur for each list of 5 neurons, our overall probability is P(5 neurons fire) = ${\binom {6}{5}}$(.3)5(.7) 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 ${\binom {6}{4}}$, giving P(4 neurons fire) = ${\binom {6}{4}}$(.3)4(.7)2 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: $P({\text{muscle contracts}})=(.3)^{6}+{\binom {6}{5}}(.3)^{5}(.7)+{\binom {6}{4}}(.3)^{4}(.7)^{2}$

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