Skip to main content

Essentials of Mathematical Probability and Statistics

John Travis

Section 6.7 Exercises

You have an urn with 10 marbles of which \(n_1 = 6\) are red and \(N_2 = 4\) are blue. You select randomly r = 3 of the marbles without replacement and let \(X\) represent the number of red marbles in your sample. With \(R\) = {0, 1, 2, 3}, determine:
  • f(x)
  • P(2 of the 3 are red) = f(2)
  • P(at most 2 of the 3 are red) = f(0) + f(1) + f(2)
Hint.
This is hypergeometric using
\begin{equation*} f(x) = \frac{\binom{6}{x} \binom{4}{3-x}}{\binom{10}{3}}. \end{equation*}
You randomly select a hand of five cards without replacement from an ordinary deck of playing cards.
  • Determine the probability that four of the five are spades.
  • Determine the probability that three of the five are face cards (ie, Jacks, Queens, Kings).
Hint.
This exercise is actually two different hypergeometric distributions: the first is the 13 spades vs the 39 other cards and the second is the 12 face cards vs the 40 other cards.
Solution.
\begin{equation*} \frac{\binom{13}{4} \binom{39}{1}}{\binom{52}{5}} \end{equation*}
\begin{equation*} \frac{\binom{12}{3} \binom{40}{2}}{\binom{52}{5}} \end{equation*}
You are picking an eleven member football starting team by picking randomly from a group with 15 seniors and 35 others. Determine:
  • P(all seniors)
  • P(exactly 6 seniors)
  • the expected number of seniors on the team
  • If your team has all seniors, explain whether someone could suggest that your decision on members was unfair
Hint.
This is a hypergeometric distribution with \(n_1 = 15, n_2 = 35, r = 11\text{.}\)
Ole Faithful geyser in Yellowstone National Park erupts every 91 minutes. You show up at some random time in the eruption cycle and your tour bus plans to stay for 25 minutes. Determine the likelihood that you will be able to see it erupt. Express your answer by giving correct formulas for \(f(x)\) and \(F(x)\) and then determine the specific answer to this question.
Hint.
This is the continuous uniform distribution over \(R\) = [0, 91].
Explain how the following situations can be modeled using either a discrete or continuous uniform distribution by identifying the space \(R\) and the corresponding \(f(x)\) for each situation.
  • The location on a prize wheel where the spun wheel will stop.
  • Given a clock with only a minute hand, the current one second interval.
  • The location on a automobile tire where the next puncture will occur.
Determine an explicit formula for \(f(x)\) and the mean and variance for an continuous uniform distribution over \(R = [-2,3] \cup [5,6] \cup [9,15]\text{.}\)
Solution.
Since you must have
\begin{equation*} \int_{x \in R} f(x) dx = 1 \end{equation*}
and since \(f(x)\) must be constant than all you must do is measure the accumulated width of the intervals in \(R\text{.}\) This is 5 + 1 + 6 = 12 and so
\begin{equation*} f(x)=\left\{\begin{matrix} \frac{1}{12}, & -2 \le x \le 3 \\ \frac{1}{12}, & 5 \le x \le 6 \\ \frac{1}{12}, & 9 \le x \le 15 \\ 0, & \text{otherwise} \end{matrix}\right. \end{equation*}
For the mean,
\begin{align*} \int_{x \in R} x \frac{1}{12} dx & = \int_{-2}^3 \frac{x}{12} dx + \int_5^6 \frac{x}{12} dx + \int_9^{15} \frac{x}{12} dx\\ & = \frac{9-4}{24} + \frac{36-25}{24} + \frac{225-81}{24}\\ & = \frac{5+11+144}{24} = \frac{160}{24} = \frac{20}{3}. \end{align*}
For the variance,
\begin{align*} \int_{x \in R} x^2 \frac{1}{12} dx - \mu^2 & = \int_{-2}^3 \frac{x^2}{12} dx + \int_5^6 \frac{x^2}{12} dx + \int_9^{15} \frac{x^2}{12} dx - \mu^2\\ & = \frac{81+8}{36} + \frac{216-125}{36} + \frac{3375-729}{36} - \big ( \frac{20}{3} \big )^2\\ & = \frac{89+91+2646}{36} - \frac{400}{9} = \frac{2826-1600}{36} \\ & = \frac{1226}{36} \approx 34.055. \end{align*}
To play the Mega Millions Louisiana Lottery consists of picking five numbers from 1 to 75 and one yellow Mega Ball number from 1 through 15. (You can play up to five different sets of numbers on each playslip but we will just assume one play per ticket to keep things straight.) Each play costs $1 and you can pay an additional $1 to apply a "multiplier" which multiplies any non-Jackpot prize by the Multiplier number (2, 3, 4, or 5) randomly selected at the time of the drawing.
On October 10, 2016 the jackpots listed were
  • Match 5 plus Mega ball = Jackpot of $49,000,000 with cash value of $32,600,000
  • Match only 5 = $1,000,000
  • Match 4 plus Mega ball = $5,000
  • Match only 4 =$500
  • Match 3 plus Mega ball = $50
  • Match only 3 = $5
  • Match 2 plus Mega ball = $5
  • Match 1 plus Mega ball = $2
  • Match only the Mega ball = $1
Verify the posted odds
  • Match 5 plus Mega ball = 1 in 258,890,850
  • Match only 5 = 1 in 18,492,204
  • Match 4 plus Mega ball = 1 in 739,688
  • Match only 4 = 1 in 52,835
  • Match 3 plus Mega ball = 1 in 10,720
  • Match only 3 = 1 in 766
  • Match 2 plus Mega ball = 1 in 473
  • Match 1 plus Mega ball = 1 in 56
  • Match only the Mega ball = 1 in 21
Determine the expected payout for each ticket purchased. Also, determine what the Jackpot would need to be in order for the game to be considered "fair" with an expected value of zero.
Solution.
Throughout these calculations, you can presume that the first five numbers are selected independently from the Mega Ball number. However, the first five numbers are selected without replacement so computing probabilities with those does not allow for independence. This part is hypergeometric with the \(n_1 = 5\) numbers you selected being the "desired" numbers and the Lottery Commission picking a subset of size r = 5 from the 75 possible numbers. So, your likelihood of matching all five would be
\begin{equation*} \frac{\binom{5}{5} \cdot \binom{70}{0}}{\binom{75}{5}} = \frac{1}{17259390}. \end{equation*}
Multiplying this by the 1 chance in 15 that you also match the Mega Ball gives
\begin{equation*} P(\text{Match 5 plus Mega Ball}) = \frac{1}{17259390} \cdot \frac{1}{15} = \frac{1}{258,890,850}. \end{equation*}
To match only 5 means you also MUST miss the Mega Ball which has probability 14/15 to give
\begin{equation*} \frac{1}{17259390} \cdot \frac{14}{15} = \frac{1}{17259390 \cdot \frac{15}{14}} \approx \frac{1}{18492204}. \end{equation*}
Continue in this manner to determine the other odds.
For the expected earnings, first determine a value function corresponding to each outcome and apply the discrete expected value process. This gives
\begin{align*} & \$32600000 \cdot \frac{1}{258,890,850} + \$1000000 \cdot \frac{1}{18,492,204} \\ & + \$5000 \cdot \frac{1}{739,688} + \$500 \cdot \frac{1}{52,835}\\ & + \$50 \cdot \frac{1}{10,720} + \$5 \cdot \frac{1}{766} \\ & + \$5 \cdot \frac{1}{473} + \$2 \cdot \frac{1}{56} + \$1 \cdot \frac{1}{21}\\ & \approx \$0.3013. \end{align*}
So, the expected payout is approximately 30 cents. Subtracting the cost of playing ($1) indicates that the average winnings per play of the Louisiana Lottery would be -70 cents. So, you would be better off to take, say, 50 cents and just give it to the local school system every time you consider playing this game rather than actually playing.
To determine the Jackpot A needed to make this a fair game means to solve the equation
\begin{align*} & A \cdot \frac{1}{258,890,850} + \$1000000 \cdot \frac{1}{18,492,204} \\ & + \$5000 \cdot \frac{1}{739,688} + \$500 \cdot \frac{1}{52,835}\\ & + \$50 \cdot \frac{1}{10,720} + \$5 \cdot \frac{1}{766} \\ & + \$5 \cdot \frac{1}{473} + \$2 \cdot \frac{1}{56} + \$1 \cdot \frac{1}{21}\\ & = 1 \end{align*}
for A.
Finally, to deal with the multiplier, note that all but the Jackpot payouts would be increased by the multiplier m where \(m \in \{1,2,3,4,5\}\text{.}\) For the cost of an extra $1 (total cost of $2 per bet) the expected payout increases as the multiplier increases but each of these decreases likelihood of winning that payout by a factor of 1/5. In general, let x = 1, 2, ..., 9 indicate the various winning options in order listed above, f(x) the corresponding probabilities listed for each option, and u(x) the listed payouts. Then the expected payout is given by
\begin{equation*} \$32600000 \cdot \frac{1}{258,890,850} + \sum_{m=1}^5 \sum_{x=2}^9 m \cdot u(x) f(x)/5 \end{equation*}
or
\begin{align*} \$32600000 \cdot \frac{1}{258,890,850} & + \sum_{m=1}^5 \frac{m}{5} \sum_{x=2}^9 u(x) f(x) \\ & = \frac{\$32600000}{258890850} + \sum_{m=1}^5 \frac{m}{5} 0.17539\\ & = 0.12592 + 3 \cdot 0.17539\\ & = 0.65209 \end{align*}
Therefore, the expect value of spending another dollar to get the multiplier effect is about -$1.35. Since this is slightly less than doubling the expected loss of 70 cents for playing without the multiplier with $1, it make more sense to bet $2 once rather than betting $1 twice. Or, you can send the extra nickel to this author of this text and call it quits.
<Prev^TopNext>
Authored in PreTeXt
Powered by Runestone