Discrete Probability Examples

Formulas Example

For a normal dice, we have the same probability of having each one of the 6 values: $\frac{1}{6}$. That's also the case in a card game, we have $\frac{1}{52}$ if we have 52 cards.

Combinatorics Example

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

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

We are introducing A = "at least one ace". It would be easier to calculate $\mathbb{P}(A) = 1 - \overline{A}$ (with $\overline{A}$=0 ace).

@ \mathbb{P}(\overline{A}) := \frac{|\overline{A}|}{|\Omega|} = \frac{C_{28}^5}{C_{32}^5} @

For $|\overline{A}|$, we are basically calculating the event of picking 5 cards among the 28 remaining cards, e.g., the deck without the aces.

Same question if Y took one ace

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

Conditional Probabilities Example

We got 3 bags with marbles:

  • B1 has $2$ white marbles and $3$ red marbles
  • B2 has $2$ green marbles and $4$ white marbles
  • B3 has $5$ black marbles and $2$ red marbles

We pick a marble in B1 and add it into B2. We repeat the process with B2. Finally, we pick a marble in B3.

What's the probability of all 3 marbles having different colors?

  • $W_i$ = "white marble picked inside B_i"
  • $A$ = "the 3 marbles are different"
  • We are also considering $R_i$ Red, $G_i$ Green, $B_i$ Black
  • There are 4 outcomes: $WGB$, $WGR$, $RGB$, $RWB$

$\mathbb{P}(A)=P(B_1V_2N_3)+P(B_1V_2R_3)+P(R_1V_2N_3)+P(R_1B_2N_3)$

  • $\mathbb{P}(W_1G_2B_3)=\frac{2*2*5}{5*7*8}=20/280$
  • $\mathbb{P}(W_1G_2R_3)=\frac{2*2*2}{5*7*8}=8/280$
  • $\mathbb{P}(R_1G_2B_3)=\frac{3*2*5}{5*7*8}=30/280$
  • $\mathbb{P}(R_1B_2B_3)=\frac{3*4*5}{5*7*8}=60/280$
  • $\mathbb{P}(A)=20/280 + 8/280 + 30/280 + 60/280 = 118/280$

Detailed calculation for $\mathbb{P}(W_1G_2N_3)=\frac{2*2*5}{5*7*8}=20/280$

  • 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 $\mathbb{P}(B_1V_2N_3)=\frac{2*2*5}{5*7*8}$

Bayes Theorem Example

We are using a test to check if the patient got the disease or not. On a sick patient, 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.

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's the probability of someone having the disease after two positive tests?

  • $D$ = "2 positives tests"
  • $P_1$ = "The first test is positive"
  • $P_2$ = "The second test is positive"
\[ \begin{split} \mathbb{P}(S|D) = \frac{\mathbb{P}(S \cap D)}{\mathbb{P}(D)}\\ =^{independence} \frac{\mathbb{P}(S) * \mathbb{P}(P_1|S) * \mathbb{P}(P_2|S)}{P(P_2) * \mathbb{P}(P_1)}\\ = \frac{(0.0005*0.96)^2}{(0.96*0.0005+0.02*(1-0.0005))^2 } \\ = \frac{0.0005*0.96^2}{0.96^2*0.0005+0.02^2*(1-0.0005)} \\ = \frac{0.0004608}{0.0008606} = 0.53544039043 \approx 0.53 \end{split} \]

Expected Value Example

We got two white marbles and three black ones in a box. Four players A, B, C, and D are picking one marble one by one. The first one that picks a white marble wins $10.

What's the expected gain for each person?

  • $\mathbb{W}_A$ = "The player A picked a white marble"
  • $\mathbb{W}_B$ = "The player B picked a white marble"
  • $\mathbb{W}_C$ = "The player C picked a white marble"
  • $\mathbb{W}_D$ = "The player D picked a white marble"

We can calculate the probabilities for each event:

  • $\mathbb{P}(W_A) = 2/5$
  • $\mathbb{P}(W_B) = \mathbb{P}(W_B|\overline{W_A}) = 2/4 * 3/5 = 6/20 = 3/10$
  • $\mathbb{P}(W_C) = \mathbb{P}(W_C|\overline{W_A \cap W_B}) = 2/3 * 3/5 * 2/4 = 12/60 = 1/5$
  • $\mathbb{P}(W_D) = \mathbb{P}(W_D|\overline{W_A \cap W_B \cap W_C}) = 1 * 3/5 * 2/4 * 1/3 = 1/10$

And the expected value for each event:

  • $\mathbb{E}[W_A] = 2/5 * 10 = 4$
  • $\mathbb{E}[W_B] = 3/10 * 10 = 3$
  • $\mathbb{E}[W_C] = 1/5 * 10 = 2$
  • $\mathbb{E}[W_D] = 1/10 * 10 = 1$

Cumulative Distribution Function Example

We are working with the function $f(x) = \frac{e^{-1}}{x!},\quad x \in \mathbb{Z}^*$.

Calculate $P(X=2)$

$\mathbb{P}(X=2) = \frac{e^{-1}}{2!}$.

Calculate $P(X < 2)$

$\mathbb{P}(X < 2) = P(X=0) + P(X=1) = \frac{e^{-1}}{0!} + \frac{e^{-1}}{1!} = e^{-1} + e^{-1} = 2 e^{-1}$