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

MATH103 April 2011

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[hide] 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

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[show] 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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[show] 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, ${\begin{aligned}{\overline {N}}=f'(q)={\frac {p}{(1-q)^{2}}}={\frac {p}{p^{2}}}={\frac {1}{p}}={\frac {7}{2}}=3.5.\end{aligned}}$