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Tutor profile: Dev O.

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Dev O.
Advanced Math Tutor for three years, Engineering at UC Berkeley
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Questions

Subject:Applied Mathematics

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Question:

Henry works for the council and is responsible for the up keep of speed cameras on roads in the area. Recently a particular camera has stop working, and Henry must go and repair it. However, to diagnose the issue and then complete the repairs, Henry must shut down the camera completely for 2 hours. The council tells Henry that the camera catches speeding drivers at a rate of 3 per hour. Henry arrives at the camera location at 8:45AM. After waiting 15 minutes, he observes no speeding drivers and decides to start his repairs. What is the probability that any speeding cars will pass the camera while it is down?

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Dev O.

In Poisson processes the distribution of the waiting times is memoryless so all you need is to compute the probability of at least one arrival in the next 2 hours. It is probably easiest to look at the Poisson process. An average of 3/h means an expected value of $$𝜇 = 6$$ over 2 hours. You then need $$𝑃(𝑋≥1)=1 − 𝑃(𝑋=0)= \dfrac{\mu^0e^{-\mu}}{0!}=1−𝑒^{-6}≈0.9975$$ If you want to go the exponential route then consider that inter-arrival times follow an exponential distribution with mean 1/3 (i.e. 20 minutes). You want to compute $$𝑃 (𝑇 < 2) = 1 − 𝑒^{−\dfrac{21}{1/3}}= 1 − 𝑒^{−6} ≈ 0.9975$$.

Subject:Discrete Math

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Question:

Suppose you have a well-shuffled standard deck of cards, (faces down). You start turning over the cards, one at a time, from the top of the deck. Let $$S$$ be the expected number of cards you need to turn over before revealing all four $$7$$'s. If $$S = \dfrac{a}{b}$$, where $$a$$ and $$b$$ are positive coprime integers, then find $$a + b$$

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Dev O.

This scenario is equivalent to finding the expected position of the last $$7$$ in a string of $$x$$'s and $$7$$'s, where the $$x$$'s represent any denomination other than a $$7$$. Now there are $$\binom{52}{4}$$ ways the four 7's can be positioned in the deck. There is one way that the string can have the last 7 in the 4th position, $$\binom{4}{3}$$ ways in which the last 7 shows up in the 5th position, and in general, $$\binom{n}{3}$$ ways in which the last 7 can show up in the $$(n + 1)$$st position for $$3 \le n \le 51$$ The expected value is then the weighted average of these individual values, i.e., $$\dfrac{1}{\binom{52}{4}}*\displaystyle\sum_{n=3}^{51} (\dbinom{n}{3}*(n + 1)) = \dfrac{4}{\binom{52}{4}}*\sum_{n=4}^{52} \dbinom{n}{4} = \dfrac{4}{\binom{52}{4}}*\dbinom{53}{5}$$ where this last step made use of the Hockey-Stick Identity. Simplifying, this last value becomes $$\dfrac{\frac{4*53!}{48!*5!}}{\frac{52!}{48!*4!}} = \dfrac{4*53}{5} = \dfrac{212}{5}$$. . Thus $$a + b = 212 + 5 = 217$$

Subject:Calculus

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Question:

See here: https://imgur.com/a/o1fQrcN

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Dev O.

I have done an explanation of the problem here: https://youtu.be/m-ga0NTth1k

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