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As­sign­ment 3

The 5-107 Lot­tery

In the 5-107 Lot­tery you choose a set of $6$ dis­tinct in­te­gers $\{x_1,\ldots,x_5,y\}$ from the set $\{1,2,3,\ldots,107\}$. $x_1,\ldots,x_5$ are called your main num­bers and $y$ is your bonus num­ber. On Fri­day night, the lot­tery ma­chine draws a uni­formly ran­dom $5$-num­ber sub­set $\{z_1,\ldots,z_5\}$ from $\{1,\ldots,107\}$. You buy one lot­tery ticket with your favourite $6$ num­bers.

Big Jack­pot

You win a Big Jack­pot if $\{x_1,\ldots,x_5\}=\{z_1,\ldots,z_5\}$. What is the prob­a­bil­ity that you win the Big Jack­pot?

So­lu­tion: The sam­ple space for this ques­tion is the set $S$ of all $\binom{107}{5}$ $5$-el­e­ment sub­sets of $\{1,\ldots,107\}$. An el­e­ment in $S$ rep­re­sents the num­bers $\{z_1,\ldots,z_5\}$ cho­sen by the ma­chine. This is a uni­form sam­ple space, so $\Pr(\omega)=1/|S|=1/\binom{107}{5}$ for each $\omega\in S$. Let $A$ be the event "you win the Big Jack­pot". Then $\Pr(A)=\Pr(\{x_1,\ldots,x_5\})=1/\binom{107}{5}$.

Grad­ing: Five marks for a cor­rect an­swer, which mainly im­plies that the size of the sam­ple space was com­puted cor­rectly. It's hard to offer part marks here since that's re­ally the only thing one needs to solve the ques­tion.

Lit­tle Jack­pot

You win a Lit­tle Jack­pot if $|\{x_1,\ldots,x_5\}\cap\{z_1,\ldots,z_5\}|=4$. What is the prob­a­bil­ity that you win a Lit­tle Jack­pot?

So­lu­tion: Let $B$ be the event "you win a Lit­tle Jack­pot" and, for each $i\in\{1,\ldots,5\}$, let $B_i$ be the event "$\{x_1,\ldots,x_5\}\cap\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_5\}\setminus\{x_i\}$. In words, $B_i$ is the event "all your num­bers matched ex­cept for $x_i$". Clearly the events $B_1,\ldots,B_5$ are pair­wise dis­joint. There­fore $\Pr(B)=\Pr(\bigcup_{i=1}^5 B_i)=\sum_{i=1}^5 \Pr(B_i)$, by the Sum Rule.

Now we just need to fig­ure out $\Pr(B_i)=|B_i|/|S|=|B_i|/\binom{107}{5}$. With­out loss of gen­er­al­ity, we focus on $\Pr(B_5)$. In words, the event $B_5$ hap­pens when $\{z_1,\ldots,z_5\}$ is a $5$-el­e­ment sub­set of $\{1,\ldots,107\}\setminus\{x_5\}$ that con­tains $\{x_1,\ldots,x_4\}$. In other words $\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_4,z_5\}$, where $z_5$ is any in­te­ger in the $102$-el­e­ment set $\{1,\ldots,107\}\setminus\{x_1,\ldots,x_5\}$. There­fore $|B_5|=\binom{102}{1}=102$. There is noth­ing spe­cial about $x_5$ in this ar­gu­ment, so $$\Pr(B_1)=\Pr(B_2)=\cdots=\Pr(B_5)=|B_5|/\binom{107}{5}=102/\binom{107}{5} \enspace .$$ Now we can fin­ish off what we started with $$\Pr(B)=\Pr\left(\bigcup_{i=1}^5 B_i\right)=\sum_{i=1}^5 \Pr(B_i)=5\cdot\Pr(B_5)=510/\binom{107}{5} \enspace .$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

Bonus Jack­pot

If you win a Lit­tle Jack­pot and $y\in\{z_1,\ldots,z_5\}$, then you win a Bonus Jack­pot. What is the prob­a­bil­ity that you win a Bonus Jack­pot?

So­lu­tion:
Let $C$ be the event "you win a Bonus Jack­pot" and, for each $i\in\{1,\ldots,5\}$, let $C_i$ be the event "$\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_5,y\}\setminus\{x_i\}$. In words, $C_i$ is the event "you win the Bonus Jack­pot be­cause all your num­bers matched ex­cept for $x_i$ and the ma­chine picked your bonus num­ber $y$". Just like the last ques­tion, $C_1,\ldots,C_5$ are pair­wise dis­joint, so $\Pr(C)=\Pr(\bigcup_{i=1}^5 C_i)=\sum_{i=1}^5 \Pr(C_i)$, by the Sum Rule. Also, like the last ques­tion, we can focus on $\Pr(C_5)$ since there is noth­ing spe­cial about $x_5$.

But now, $C_5$ is easy to de­scribe. It hap­pens pre­cisely when $\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_4,y\}$. There­fore, $|C_5|=1$ and $\Pr(C_5)=1/\binom{107}{5}$. There­fore, $$\Pr(C)=\Pr\left(\bigcup_{i=1}^5 C_i\right)=\sum_{i=1}^5 \Pr(C_i)=5\cdot\Pr(C_5)=5/\binom{107}{5} \enspace .$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

Ques­tion 2

I have a card game with $100$ cards. For each $i\in\{1,\ldots,100\}$ there is a card with the num­ber $i$ printed on it. I take 5 cards from the deck and get: $\{56, 55, 46, 1, 33\}$. You take a uni­formly ran­dom $5$-el­e­ment sub­set from the re­main­ing cards.

High­est card wins

What is the prob­a­bil­ity that my high­est card (56) is higher than your high­est card?

So­lu­tion: The sam­ple space $S$ here rep­re­sents the $5$ cards in your hand, and it's a uni­formly ran­dom $5$-el­e­ment sub­set of the $95$-el­e­ment set $\{1,\ldots,100\}\setminus \{56, 55, 46, 1, 33\}$. There­fore $|S|=\binom{95}{5}$ and $\Pr(\omega)=1/|S|=1/\binom{95}{5}$ for each $\omega\in S$.

Let $A$ be the event "your high­est card is lower than $56$". Then $A$ oc­curs if and only if your cards are a $5$-el­e­ment sub­set of the $51$-card set $\{1,\ldots,54\}\setminus\{1,46,33\}$. The num­ber of such sub­sets is $|A|=\binom{51}{5}$, so $$\Pr(A)=\frac{|A|}{|S|}=\frac{\binom{51}{5}}{\binom{95}{5}} = \frac{55930}{1107249} \approx 0.040542612329723865 \enspace .$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

High­est ver­sus low­est

What is the prob­a­bil­ity that my high­est card (56) is lower than your low­est card?

So­lu­tion: This is sim­i­lar to the last ques­tion. Let $B$ be the event "your low­est card is higher than $56$". Then $B$ con­tains every $5$-el­e­ment sub­set of $44$-el­e­ment set $\{57,\ldots,100\}$, so $|B|=\binom{44}{5}$ and $$\Pr(B)=\frac{\binom{44}{5}}{\binom{95}{5}} = \frac{155144}{8277217} \approx 0.018743497965560164 \enspace .$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

Sec­ond-high­est ver­sus sec­ond-high­est

What is the prob­a­bil­ity that my sec­ond-high­est card ($55$) is higher than your sec­ond high­est card?

So­lu­tion: Let $C$ be the event "your sec­ond high­est card is lower than $55$". We break $C$ up into two dis­joint events, so that we can use hte Sum Rule:

1. $C_0$ is the event "all your cards are lower than $55$". This is ac­tu­ally the same as the event $A$ in the first part of the prob­lem, so $\Pr(C_0)=\Pr(A)=\binom{51}{5}/\binom{100}{5}$.
2. $C_1$ is the event "one of your cards is higher than $55$ and the other $4$ are lower than $55$". We can com­pute $|C_1|$ using the Prod­uct Rule. We can choose one num­ber $z_1$ from the set $44$-el­e­ment set $\{57,\ldots,100\}$ and then choose $z_2,\ldots,z_5$ to be a $4$-el­e­ment sub­set of the $51$-el­e­ment set $\{1,2,\ldots,54\}\setminus\{1,33,46\}$. There­fore, $|C_1|=44\cdot\binom{51}{4}$.

There­fore $$\begin{align*} \Pr(C) & = \Pr(C_0\cup C_1) \\ & = \Pr(C_0)+\Pr(C_1) & \text{(Sum Rule)} \\ & = \frac{\binom{51}{5}}{\binom{100}{5}} + \frac{|C_1|}{|S|} \\ & = \frac{\binom{51}{5}}{\binom{95}{5}} + \frac{44\cdot\binom{51}{4}}{\binom{95}{5}} \\ & = \frac{1906380}{8277217} & \text{(Using Python)} \\ & \approx 0.2303165423837505 & \text{(Using a calculator)} \end{align*}$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

Na­tional-Pub­lic redux

NPR-1

You play a game where you roll an $8$-sided die $8$ times and you win if you roll $8$ at least once. What is the prob­a­bil­ity that you win?

So­lu­tion: For this ques­tion, the sam­ple space $S:=\{1,\ldots,8\}^8$, $|S|=8^8$ and this is a uni­form prob­a­bil­ity space, so $\Pr(\omega)=1/|S|=1/8^8$ for each $\omega\in S$. Let $E$ be the event "I win the game". The eas­i­est thing to do here is to com­pute $\Pr(\overline{E})$ and use the Com­ple­ment Rule. If we never roll an $8$, then each of the $8$ roles is in $\{1,\ldots,7\}$, so $\overline{E}=\{1,\ldots,7\}^8$ and $\Pr(\overline{E})=|\overline{E}|/|S|=(7/8)^8$. There­fore $$\Pr(E)=1-\Pr(\overline{E}) = 1-\frac{|\overline{E}|}{|S|}= 1-\left(\frac{7}{8}\right)^8 = \frac{11012415}{16777216}\approx 0.6563910841941833$$

NPR-2

You play a game where you roll an $8$-sided die $16$ times and you win if you roll $8$ at least twice. What is the prob­a­bil­ity that you win?

So­lu­tion: For this ques­tion, the sam­ple space $S:=\{1,\ldots,8\}^{16}$, $|S|=8^{16}$ and this is a uni­form prob­a­bil­ity space, so $\Pr(\omega)=1/|S|=1/8^{16}$ for each $\omega\in S$. Let $E$ be the event "I win the game". The eas­i­est thing to do here is to com­pute $\Pr(\overline{E})$ and use the Com­ple­ment Rule. To com­pute $\Pr(\overline{E})$, we will par­ti­tion $\overline{E}$ into two dis­joint events and use the Sum Rule:

1. Let $\overline{E}_0$ be the event "I roll 8 zero times". Then $|\overline{E}_0|=7^{16}$.
2. Let $\overline{E}_1$ be the event "I roll 8 ex­actly once". We can com­pute $|\overline{E}_1|$ using the Prod­uct Rule: First choose which of the $16$ rolls will be an $8$ and then a value in $\{1,\ldots,7\}$ for each of the $15$ other rolls. There­fore, $|\overline{E}_1|=16\cdot 7^{15}$.

We fin­ish with $$\begin{align*} \Pr(E) & =1-\Pr(\overline{E}) & \text{(Complement Rule)} \\ & = 1-(\Pr(\overline{E}_0) + \Pr(\overline{E}_1)) & \text{(Sum Rule)} \\ & = 1-\left(\frac{|\overline{E}_0| + |\overline{E}_1|}{|S|}\right) & \text{(Uniform probability space)} \\ & = 1-\left(\frac{7^{16} + 16\cdot 7^{15}}{8^{16}}\right) \\ & = 1-\left(\frac{172281061981967}{281474976710656}\right) & \text{(Python)} \\ & \approx 0.6120652855016111 \end{align*}$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

A Drink­ing Game

Michiel and Pat are on the bal­cony with a cooler that has 20 bot­tles of beer in it: ten bot­tles of lager and ten bot­tles of IPA. They play a game where Pat takes a ran­dom beer from the cooler, drinks it, and throws the bot­tle off the bal­cony, then Michiel does the same. They do this for two rounds, until the neigh­bours call the po­lice.

First IPA Wins (Pat)

In the game of First IPA wins, the per­son who drinks the first bot­tle of IPA is the win­ner (this game can end in a draw if no one drinks an IPA). What is the prob­a­bil­ity that Pat wins this game?

So­lu­tion: There are dif­fer­ent ways to setup the sam­ple space here, but one of the eas­i­est is to de­fine the bot­tle set $B=\{I_1,\ldots,I_{10},L_1,\ldots,L_{10}\}$. Then play­ing the game cor­re­sponds to choos­ing an or­dered $4$-el­e­ment sub­set of $B$ which rep­re­sents the order in which four bot­tles of beer are drank (drunk?). The size of the re­sult­ing sam­ple space $S$ is then $20\cdot 19\cdot 18\cdot 17=20!/16!$. (We showed this when prov­ing that $\binom{n}{k}=n!/(k!(n-k)!)$.) This is a uni­form sam­ple space, which will make things eas­ier.

Let $P$ be the event "Pat wins the game of First IPA Wins". Par­ti­tion $P$ into the two dis­joint events:

1. $P_1$ is the event "Pat wins in the first round". Then $|P_1|=10\cdot 19\times 18\times 17=\tfrac{1}{2}|S|$ since $P_1$ oc­curs pre­cisely when Pat chooses one of the $10$ IPA bot­tles in the first round, after which any­thing goes.
2. $P_2$ is the event "Pat wins in the sec­ond round". Then $|P_2|=10\cdot 9\cdot 10\cdot 17$ since $P_2$ hap­pens when Pat drinks one of the 10 lagers, then Michiel drinks one of the 9 re­main­ing lagers, then Pat drinks one of the 10 IPA, after which any­thing goes.

Since $P_1$ and $P_2$ are dis­joint and $P=P_1\cup P_2$, $$\begin{align*} \Pr(P) & =\Pr(P_1\cup P_2) \\ & = \frac{|P_1\cup P_2|}{|S|} \\ & = \frac{|P_1|+|P_2|}{|S|} \\ & = 1/2 + \frac{10\cdot 9\cdot 10\cdot 17}{20\cdot 19\cdot 18\cdot 17} \\ & = 12/19 \approx 0.6315789 \end{align*}$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

First IPA wins (Michiel)

What is the prob­a­bil­ity that Michiel wins the game of First IPA wins?

So­lu­tion: Let $M$ be the event "Michiel wins the game of First IPA Wins". We par­ti­tion $M$ into two events so that we can use the Sum Rule.

1. $M_1$ is the event "Michiel wins in the first round". Then $|M_1|=10\cdot 10\cdot 18\cdot 17$ since $M_1$ oc­curs when Pat drinks one of the 10 lagers, then Michiel drinks one of the ten IPA, after which any­thing goes.
2. $M_2$ is the event "Michiel wins in the sec­ond round". Then $|M_2|=10\cdot 9\cdot 8\cdot 10$ since $M_2$ oc­curs when Pat drinks one of the 10 lagers then Michiel drinks one of the $9$ re­main­ing lagers, then Pat drinks one of the $8$ re­main­ing lagers, then Michiel drinks one of the $10$ IPA.

Since $M_1$ and $M_2$ are dis­joint and $M=M_1\cup M_2$, we pro­ceed as above to get $$\Pr(M)=\frac{|M_1|+|M_2|}{|S|} = \frac{10\cdot 10\cdot 18\cdot 17+10\cdot 9\cdot 8\cdot 10}{20\cdot 19\cdot 18\cdot 17} = 105/323 \approx 0.32507$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error. For this, and the next cou­ple of ques­tions, stu­dents may have de­fined (even im­plic­itly) a dif­fer­ent sam­ple space in which the game ends as soon as there is a win­ner. Then they use a ``de­ci­sion-tree'' ar­gu­ment in which they de­rive that $\Pr(P_1)=10/20$ since with prob­a­bil­ity $10/20$ the first beer Pat drinks is an IPA. This kind of ar­gu­ment is ok too, pro­vided that they get the cor­rect an­swer.

Sec­ond IPA Wins (Pat)

In the game of Sec­ond IPA wins, the per­son who drinks the sec­ond bot­tle of IPA is the win­ner. What is the prob­a­bil­ity that Pat wins this game?

So­lu­tion:

Let $P$ be the event "Pat wins the game of Sec­ond IPA Wins". We will use the Sum Rule again, with the fol­low­ing two events:

1. $P_1$ is the event "Pat drinks the first IPA and the sec­ond IPA". Then $|P_1|=10\cdot 10\cdot 9\cdot 17$ since $A_1$ oc­curs when Pat drinks one of the 10 IPA, then Michiel drinks one of the ten lagers, then Pat drinks one of the 9 re­main­ing IPAs, then any­thing goes.
2. $P_2$ is the event "Michiel drinks the first IPA and Pat drinks the sec­ond IPA". Then $|P_2|=10\cdot 10\cdot 9\cdot 17$ since $A_2$ oc­curs when Pat drinks one of the 10 lagers, then Michiel drinks one of the 19 IPAs then Pat drinks one of the 9 re­main­ing IPAs then any­thing goes.

Then, $$\Pr(P) = \Pr(P_1\cup P_2) =\frac{|P_1\cup P_2|}{|S|} =\frac{10\cdot 10\cdot 9\cdot 17+10\cdot 10\cdot 9\cdot 17}{20\cdot 19\cdot 18\cdot 17} = \frac{5}{19} \approx 0.2631578947368421 \enspace .$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error.

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error. A de­ci­sion-tree style of ar­gu­ment is also ac­cept­able here, pro­vided that it gives the cor­rect an­swer.

Sec­ond IPA Wins (Michiel)

What is the prob­a­bil­ity that Michiel wins the game of Sec­ond IPA Wins?

So­lu­tion: Let $M$ be the event "Michiel wins the game of Sec­ond IPA Wins". We will use the Sum Rule again, with the fol­low­ing two events:

1. $M_0$ is the event "Michiel wins in the first round". Then $|M_0|=10\cdot 9\cdot 18\cdot 17$ since $M_0$ oc­curs when Pat drinks one of the $10$ IPA, then Michiel drinks one of the $9$ re­main­ing IPA, after which any­thing goes.
2. $M_1$ is the event "Michiel drinks the first IPA and the sec­ond IPA". Then $|M_1|=10\cdot 10\cdot 9\cdot 9$.
3. $M_2$ is the event "Pat drinks the first IPA and Michiel drinks the sec­ond IPA". We can fur­ther par­ti­tion $M_2$ into two events $M_{2,1}$ and $M_{2,2}$ de­pend­ing on whether Pat drinks the IPA in the first or sec­ond round.
4. $|M_{2,1}|=10\cdot 10\cdot 9\cdot 9$ since $M_{2,1}$ oc­curs when Pat drinks one of the 10 IPAs, then Michiel drinks one of the 10 lagers, then Pat drinks one of the 9 re­main­ing lagers, then Michiel drinks one of the 9 re­main­ing IPAs.
5. $|M_{2,2}|=10\cdot 9\cdot 10\cdot 9$ since $M_{2,2}$ oc­curs when Pat drinks one of the 10 lagers, then Michiel drinks one of the $9$ re­main­ing lagers, then Pat drinks one of the 10 IPAs, then Michiel drinks one of the 9 re­main­ing IPAs.

(No­tice that $|M_1|=|M_{2,1}|=|M_{2,2}|$.) Then $$\begin{align*} \Pr(M) & = \Pr(M_0\cup M_1\cup M_{2,1}\cup M_{2,2}) \\ & = \frac{|M_0\cup M_1\cup M_{2,1}\cup M_{2,2}|}{|S|} \\ & = \frac{|M_0|+|M_1|+|M_{2,1}|+|M_{2,2}|}{|S|} & \text{(Sum Rule)} \\ & = \frac{10\cdot 9\cdot 18\cdot 17 + 3\cdot 10\cdot 10\cdot 9\cdot 9}{20\cdot 19\cdot 18\cdot 17} \\ & = 144/323 \approx 0.4458204334365325 \\ \end{align*}$$

Grad­ing: Five marks for a com­plete and cor­rect an­swer, with jus­ti­fi­ca­tion. You may give par­tial marks if the rea­son­ing was cor­rect but they made some cal­cu­la­tion error. A de­ci­sion-tree style of ar­gu­ment is also ac­cept­able here, pro­vided that it gives the cor­rect an­swer.

Are Most Horses Red?

I im­ported a state of the art gashapon ma­chine from Japan that con­tains $1000$ cap­sules, each with a toy horse in it. When I put a dol­lar into the ma­chine it gives me a ran­dom cap­sule that I can open and check the colour of the horse in­side it. The seller claims that half cap­sules con­tain a red horse and half the cap­sules con­tain a brown horse, but I've read re­views from peo­ple com­plain­ing that their ma­chines con­tain 900 red horses and only 100 brown horses.

This leads to two hy­pothe­ses:

1. $H_0$: My ma­chine con­tains 500 red horses and 500 brown horses.
2. $H_1$: My ma­chine con­tains 900 red horses and 100 brown horses.

For any ex­per­i­ment I do, the hy­poth­e­sis $H_0$ de­fines a prob­a­bil­ity func­tion $\Pr_0$ and the hy­poth­e­sis $H_1$ de­fines a dif­fer­ent prob­a­bil­ity func­tion $\Pr_1$.

I only have $\$3$ to fig­ure out which of these hy­pothe­ses is more likely.

A dumb test

Sup­pose I buy three cap­sules from the ma­chine. Let $A$ be the event "the three cap­sules I bought con­tain more red horses than brown horses". Com­pute $\Pr_0(A)$ and $\Pr_1(A)$.

So­lu­tion: The event $A$ is equiv­a­lent to "the three cap­sules I bought con­tain $2$ or $3$ red horses".

The com­ple­men­tary event $\overline{A}$ is "the three cap­sules I bought con­tain $2$ or $3$ brown horses". Under hy­poth­e­sis 0, there is com­plete sym­me­try be­tween red and brown, so $\Pr_0(A)=\Pr_0(\overline{A})$. Since $\Pr_0(A)+\Pr_0(\overline{A})=1$, this means $\Pr_0(A)=\Pr_0(\overline{A})=1/2$.

The analy­sis under hy­poth­e­sis $H_1$ takes a bit more work. Split $A$ into the two events $A_2$ and $A_3$ de­pend­ing on whether I get $2$ or $3$ red horses, re­spec­tively. Then $$\Pr_1(A_2)=\frac{3\times 900\times 899\times 100}{1000\times 999\times 998}$$ and $$\Pr_1(A_3) = \frac{900\times 899\times 898}{1000\times 999\times 998}$$ so $$\Pr_1(A)=\Pr_1(A_2) + \Pr_1(A_3) = \frac{3\times 900\times 899\times 100+900\times 899\times 898}{1000\times 999\times 998} = \frac{538501}{553890} \approx 0.9722165050822366$$

A bet­ter test

De­scribe and an­a­lyze an ex­per­i­ment I can run that costs only $\$3$ and has the fol­low­ing prop­er­ties:

• The re­sult of the ex­per­i­ment is a guess about whether $H_0$ or $H_1$ is true.
• If $H_0$ is true, the ex­per­i­ment guesses $H_0$ with prob­a­bil­ity greater than $1/2$.
• If $H_1$ is true, the ex­per­i­ment guesses $H_1$ with prob­a­bil­ity greater than $1/2$.

So­lu­tion: There's re­ally only one rea­son­able test. Buy three cap­sules. If they all con­tain red horses, guess $H_1$, oth­er­wise guess $H_0$. In the lan­guage of the pre­vi­ous an­swer, if event $A_3$ oc­curs then guess $H_1$, oth­er­wise guess $H_0$.

If $H_1$ is cor­rect, then we're run­ning the ex­per­i­ment with prob­a­bil­ity func­tion $\Pr_1$, and we al­ready know: $$\Pr_1(A_3) = \frac{900\times 899\times 898}{1000\times 999\times 998} = \frac{403651}{553890} \approx 0.7287566123237466 > 1/2 \enspace .$$

If $H_0$ is cor­rect, then we're run­ning the ex­per­i­ment with prob­a­bil­ity func­tion $\Pr_0$ and we get $$\Pr_0(A_3) = \frac{500\times 499\times 498}{1000\times 999\times 998} = \frac{83}{666} \approx 0.12462462462462462 < 1/2 \enspace .$$ so $\Pr_0(\overline{A}_3)=1-\Pr_0(A_3)>1/2$.

Grad­ing: Full marks if they de­scribe and an­a­lyze the test given in the sam­ple so­lu­tion, show­ing that the prob­a­bil­ity that the test gives the cor­rect an­swer under hy­poth­e­sis 0 is greater than $1/2$ and the prob­a­bil­ity that the test gives the cor­rect an­swer under hy­poth­e­sis $1$ is also greater than $1/2$. Note: Some thrifty stu­dents have no­ticed that there is an even cheaper test: Buy two horses: If they're both red, guess $H_1$, oth­er­wise guess $H_0$. In this case, the prob­a­bil­ity of being cor­rect under hy­poth­e­sis $H_0$ is $1-(500\cdot 499)/(1000\cdot999)\approx 3/4$ and the prob­a­bil­ity of being cor­rect under hy­poth­e­sis $H_1$ is $(900\cdot899) / (1000\cdot 999)\approx 81/100$. Any stu­dent with this an­swer should also re­ceive full marks.

Three events

Let $A,B,C\subseteq S$ be three events in a prob­a­bil­ity space $(S,\Pr)$ where

1. $S=A\cup B$,
2. $\Pr(A)=\Pr(B)=2/3$,
3. $\Pr(C)=2/5$,
4. $\Pr(A\cap C)=\Pr(B\cap C)=1/3$, and
5. $\Pr(A\cap B\cap C)=1/6$.

No­tice: This ques­tion is bro­ken be­cause
$$\begin{align*} \Pr(C) & \ge \Pr(C\cap (A\cup B)) & \text{(since $C\cap (A\cup B)\subseteq C$)}\\ & = \Pr((C\cap A)\cup (C\cap B)) \\ & = \Pr(A\cap C)+\Pr(B\cap C)-\Pr(A\cap B\cap C) & \text{(principle of inclusion-exclusion)}\\ & = 1/3+1/3-1/6 \text{(given in the question)}\\ & = 1/2 \\ & > \Pr(C) & \text{(given in the question)} . \end{align*}$$

All stu­dents should re­ceive $5/5$ for each of the two parts of this ques­tion, even if they didn't com­plete them. That is, they should re­ceive 10 free marks.

$A$ ver­sus $C$ and $B$ ver­sus $C$

Are the events $A$ and $C$ in­de­pen­dent? Are the events $B$ and $C$ in­de­pen­dent?

$(A\cup B)$ ver­sus $C$

Are the events $(A\cup B)$ and $C$ in­de­pen­dent?