Science:Math Exam Resources/Courses/MATH103/April 2011/Question 01 (c)

MATH103 April 2011

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Question 01 (c)
A dice is manipulated such that the chance of throwing a 6 is twice as likely as throwing any of the other numbers (1-5). What is the expected (average) number of throws required to get a 6? Circle the correct answer. 3 3.5 4 6 7

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
Note that each roll of the dice can be considered as a Bernoulli trial, where rolling a 6 is "success". What is the probability of success in one trial?

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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. It's probably the case that you've seen a general formula in Math 103 for this type of scenario and you are just being asked to recall it. The correct answer is 3.5. In any case, here is a full derivation, which might be useful for your review. This is a sequence of Bernoulli trials. We'll call rolling a 6 "success" and any other number "failure". In a single roll, since 6 is twice as likely as all other numbers, the probability of rolling a 6 must be 2/7 and the probability of rolling any other number must be 1/7. That is, the probability of success in one trial is $p=2/7$ and the probability of failure is $q=5/7$. Let $N$ be the number of rolls until a 6 is obtained. For example, the probability that $N=1$ is simply equal to the probability of success in one trial, $\displaystyle P(N=1)=p.$ $N=2$ if and only the first trial resulted in failure and the second trial resulted in success. Hence, $\displaystyle P(N=2)=qp.$ In general, $N=k$ if and only if the first $k-1$ trials resulted in failure and the $k^{th}$ trial resulted in success. Hence, $\displaystyle P(N=k)=q^{k-1}p.$ This is the probability distribution for the random variable $N$. Our goal is to compute the expected number of throws required to get a 6, which is precisely the mean value of $N$, ${\begin{aligned}{\overline {N}}=\sum _{k=1}^{\infty }kP(N=k)=\sum _{k=1}^{\infty }kq^{k-1}p.\end{aligned}}$ We can compute the value of this sum by making the clever observation that it appears to be the derivative of a geometric series. For any $x$ in the interval $[0,1),$ define ${\begin{aligned}f(x)=\sum _{k=0}^{\infty }x^{k}p.\end{aligned}}$ Differentiating term-by-term, we see that ${\begin{aligned}f'(x)=\sum _{k=1}^{\infty }kx^{k-1}p.\end{aligned}}$ Hence, ${\overline {N}}=f'(q).$ However, $f(x)$ is simply a geometric series; the result is ${\begin{aligned}f(x)={\frac {p}{1-x}}.\end{aligned}}$ Therefore, ${\begin{aligned}f'(x)={\frac {p}{(1-x)^{2}}}.\end{aligned}}$ Hence, ${\begin{aligned}{\overline {N}}=f'(q)={\frac {p}{(1-q)^{2}}}={\frac {p}{p^{2}}}={\frac {1}{p}}={\frac {7}{2}}=3.5.\end{aligned}}$

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