# Exercises ¶

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This is a batch of exercises on discrete probabilities (probabilities, counting, conditional probability, Bayes's formula, expected value, variance).

Exercise 1 (Counting / Conditional probability)

Two players X and Y are drawing five cards in a 32 cards deck.

What's the probability of X having at least one ace?

We are introducing A = "at least one ace" but we would have a hard time evaluating this so we will use $\mathbb{P}(A) = 1 - \overline{A}$ (with $\overline{A}$=0 ace) and @ \mathbb{P}(\overline{A}) := \frac{|\overline{A}|}{|\Omega|} = \frac{C_{28}^5}{C_{32}^5} @ We could explain that by the fact that $|\overline{A}|$ is the same as picking 5 cards in a deck after we removed the 4 aces (32-4=28).

Now Y got one ace. What's the probability of X having at least one?

Now since Y took 5 cards including 1 ace, we are simply removing them from our deck and doing the same thing. @ \mathbb{P}(\overline{A}) = \frac{C_{25}^5}{C_{28}^5} @

Exercise 2 (Conditional probability)

We got 3 bags, B1, B2, and B3. We got two white marbles (=billes) and 3 red ones. Inside B2, we got 2 green marbles and 4 white ones. Inside B3, we got 5 black marbles and 2 red ones.

We are picking a marble in B1, and add it into B2. Then, we are picking one B2 to B3. And finally, we are picking one into B3. What's the probability that the 3 marbles we picked have different colors?

1. random variables
• $W_i$ = "white marble picked inside B_i"
• We are also considering $R_i$ Red, $G_i$ Green, $B_i$ Black
2. Our goal
• A = "the 3 marbles are different"
• There are 4 cases: $WGB$, $WGR$, $RGB$, $RWB$
• $\mathbb{P}(A)=P(B_1V_2N_3)+P(B_1V_2R_3)+P(R_1V_2N_3)+P(R_1B_2N_3)$
3. Solving
• $\mathbb{P}(W_1G_2B_3)=\frac{2*2*5}{5*7*8}=20/280$
• $\mathbb{P}(W_1G_2R_3)=\frac{2*2*2}{5*7*8}=8/280$
• $\mathbb{P}(R_1G_2B_3)=\frac{3*2*5}{5*7*8}=30/280$
• $\mathbb{P}(R_1B_2B_3)=\frac{3*4*5}{5*7*8}=60/280$
• $\mathbb{P}(A)=20/280 + 8/280 + 30/280 + 60/280 = 118/280$
4. Explanation
• how did we got $\mathbb{P}(W_1G_2N_3)=\frac{2*2*5}{5*7*8}=20/280$?
• we got 5 marbles, 2 are white: $\mathbb{P}(W_1)=2/5$
• we got 6(+1) marbles, 2 are green: $\mathbb{P}(G_2|W_1)=2/7$
• we got 7(+1) marbles, 5 are black: $\mathbb{P}(B_3|W_1 \cap G_2)=5/8$
• we know: $\mathbb{P}(W_1G_2B_3)=\mathbb{P}(W_1)*\mathbb{P}(G_2|W_1)*\mathbb{P}(B_3|W_1 \cap G_2)$
• so we have $\mathbb{P}(B_1V_2N_3)=\frac{2*2*5}{5*7*8}$

Exercise 3 (Bayes' theorem)

We are using a test to check if the patient got the disease or not. If got the decease then the test is positive 96% of the time. The test is a false-positive in 2% of the cases. $0.05%$ of the patients got the decease.

1. What's the probability of someone having a positive test to have the disease?
• S = "The patient is sick"
• P = "The patient got a positive test"
• $\mathbb{P}(P|S) = 0.96$
• $\mathbb{P}(P|\overline{S}) = 0.02$
• $\mathbb{P}(S) = 0.0005$
• $\mathbb{P}(S|P) = ???$
$\begin{split} \mathbb{P}(S|P) = \frac{\mathbb{P}(S \cap P)}{\mathbb{P}(P)} =^{bayes} \frac{\mathbb{P}(S) * \mathbb{P}(P|S)}{\mathbb{P}(P)} \\ =^\text{law of total probability} \frac{\mathbb{P}(S) * \mathbb{P}(P|S)}{\mathbb{P}(P|S) * \mathbb{P}(S) + \mathbb{P}(P|\overline{S}) * \mathbb{P}(\overline{S})} \\ = \frac{0.0005*0.96}{0.96*0.0005+0.02*(1-0.0005)} = 0.02344894968 \approx 0.023\end{split}$

What we call "Law of total probability"/"Formule des probabilités totales" is a tree, considering the 2 cases (here $S$ and $\overline{S}$).

1. What's the probability of someone having the disease after two positives tests?
• $D$ = "2 positives tests"
• $P_1$ = "The first test is positive"
• $P_2$ = "The second test is positive"
$\begin{split} \mathbb{P}(S|D) = \frac{\mathbb{P}(S \cap D)}{\mathbb{P}(D)} =^{independence} \frac{\mathbb{P}(S) * \mathbb{P}(P_1|S) * \mathbb{P}(P_2|S)}{P(P_2) * \mathbb{P}(P_1)}\\ = \frac{(0.0005*0.96)^2}{(0.96*0.0005+0.02*(1-0.0005))^2 } = \frac{0.0005*0.96^2}{0.96^2*0.0005+0.02^2*(1-0.0005)} \\ = \frac{0.0004608}{0.0008606} = 0.53544039043 \approx 0.53\end{split}$

Exercise 4 (expected value)

We got 2 white marbles and 3 black ones in a box. 4 players (A, B, C, and D) taking in this order a marble. The first one taking a white marble win $10. What's the expected gain for each person? We will introduce the random variables •$\mathbb{W}_A$= "The player A got a white marble" • ... So we got •$\mathbb{P}(W_A) = 2/5$•$\mathbb{P}(W_B) = \mathbb{P}(W_B|\overline{W_A}) = 2/4 * 3/5 = 6/20 = 3/10$•$\mathbb{P}(W_C) = \mathbb{P}(W_C|\overline{W_A \cap W_B}) = 2/3 * 3/5 * 2/4 = 12/60 = 1/5$•$\mathbb{P}(W_D) = \mathbb{P}(W_D|\overline{W_A \cap W_B \cap W_C}) = 1 * 3/5 * 2/4 * 1/3 = 1/10$And •$\mathbb{E}[W_A] = 2/5 * 10 = 4$•$\mathbb{E}[W_B] = 3/10 * 10 = 3$•$\mathbb{E}[W_C] = 1/5 * 10 = 2$•$\mathbb{E}[W_D] = 1/10 * 10 = 1$Exercise 5 (Distributions) We are working with the density function$f(x) = \frac{e^{-1}}{x!},\hspace{0.5cm} x = 0,1, 2...$. 1.$P(X=2)$The answer is$\mathbb{P}(X=2) = \frac{e^{-1}}{2!}$. 1.$P(X < 2)$The answer is$\mathbb{P}(X < 2) = P(X=0) + P(X=1) = \frac{e^{-1}}{0!} + \frac{e^{-1}}{1!} = e^{-1} + e^{-1} = 2 e^{-1}$. 1. Demonstrate that$e^{-1}$is the constant making$\frac{c}{x!}$a mass function? $\begin{split} \text{mass function} := \sum_{k=0}^{+\infty} f(k) = 1 \\ \Leftrightarrow \sum_{k=0}^{+\infty} \frac{c}{k!} = 1 \Leftrightarrow c * \sum_{k=0}^{+\infty} \frac{1}{k!} = 1 \Leftrightarrow^{because\ 1^k = 1} c * \sum_{k=0}^{+\infty} \frac{1^k}{k!} = 1\\ \Leftrightarrow^\text{exponential series} c * e = 1 \Leftrightarrow c = \frac{1}{e} \Leftrightarrow c = e^{-1}\end{split}$ We also need to check that the values are$\ge 0$so that$\frac{e^{-1}}{x!} \ge 0$. Since$x! \ge 0$(because x \in R^*) and$e^{-1} \ge 0\$ then the function is a mass function.