Probabilistic Methods — Interactive Solver

Task 1 — Coin Tossing

A fair coin is tossed once, then twice, then three times. Describe the sample space for each experiment and give its size. What is an elementary outcome?

One toss

A single toss yields heads (H) or tails (T). Each is an elementary outcome.

\Omega_1 = \{H,\, T\}, \quad |\Omega_1| = 2

Two tosses

Each outcome is an ordered pair of results from the first and second toss.

\Omega_2 = \{HH,\, HT,\, TH,\, TT\}, \quad |\Omega_2| = 4

Three tosses

Each outcome is an ordered triple. Since each toss has 2 outcomes, by the multiplication rule there are $2^3$ total outcomes.

|\Omega_3| = 2^3 = 8

Full sample space for three tosses

Listing all eight triples explicitly:

\Omega_3 = \{HHH,\, HHT,\, HTH,\, HTT,\, THH,\, THT,\, TTH,\, TTT\}

General pattern

For $n$ tosses the sample space has $2^n$ elements, since tosses are independent and each doubles the count.

|\Omega_n| = 2^n

Tree diagram (two tosses)

Each toss branches into H or T; the four equally likely leaves form $\Omega_2$ .

Two coin tosses

Answer

|\Omega_1| = 2,\quad |\Omega_2| = 4,\quad |\Omega_3| = 8

Explanation tip: An elementary outcome is one complete ordered sequence of toss results; the space size doubles with each extra toss because tosses are independent and each has exactly two outcomes.

Task 2 — Rolling a Die

A fair six-sided die is rolled once, then twice, then three times. Describe the sample space for each experiment and give its size.

One roll

A single roll of a standard die shows one of six faces.

\Omega_1 = \{1,\, 2,\, 3,\, 4,\, 5,\, 6\}, \quad |\Omega_1| = 6

Two rolls

Each outcome is an ordered pair $(i, j)$ where $i, j \in \{1,\ldots,6\}$ . By the multiplication rule:

\Omega_2 = \{(i,j) : i,j \in \{1,2,3,4,5,6\}\}, \quad |\Omega_2| = 6 \times 6 = 36

Three rolls

Each outcome is an ordered triple $(i, j, k)$ . Applying the multiplication rule again:

|\Omega_3| = 6^3 = 216

General pattern

For $n$ rolls of a $k$ -sided die the sample space has $k^n$ elements.

|\Omega_n| = 6^n

Sample-space size chart

Each extra die multiplies the count by 6.

Sample-space size grows as 6^k

Answer

|\Omega_1| = 6,\quad |\Omega_2| = 36,\quad |\Omega_3| = 216

Explanation tip: Each roll adds an independent factor of 6 to the count; the sample space for

n

rolls consists of all ordered

n

-tuples from

\{1,\ldots,6\}

, giving

6^n

equally likely elementary outcomes.

Task 3 — Drawing Cards

A standard deck has 52 cards. (a) One card is drawn: describe the sample space. (b) Two cards are drawn with replacement: find the sample space size. (c) Two cards are drawn without replacement: find the sample space size.

a)(a) One card drawn at random.
b)(b) Two cards drawn with replacement (card is returned before second draw).
c)(c) Two cards drawn without replacement.

Part (a) — one card

The sample space consists of all 52 individual cards in the deck.

\Omega_1 = \{c : c \text{ is a card in the deck}\}, \quad |\Omega_1| = 52

Part (b) — two draws with replacement

After recording the first card it is returned, so the second draw also has 52 choices. Each outcome is an ordered pair.

|\Omega_2^{\text{with}}| = 52 \times 52 = 52^2 = 2704

Part (c) — two draws without replacement

The first draw has 52 choices; since the card is NOT returned, the second draw has only 51 choices. Order matters (we record which card came first).

|\Omega_2^{\text{without}}| = 52 \times 51 = 2652

Comparison

Without replacement the space is smaller because repeated outcomes (drawing the same card twice) are impossible.

2652 = 2704 - 52

Counting two-card draws

Replacement keeps all 52 cards available for the second draw; without replacement one card is gone.

Counting two-card draws

Answer

|\Omega_1| = 52,\quad |\Omega_2^{\text{with}}| = 2704,\quad |\Omega_2^{\text{without}}| = 2652

Explanation tip: With replacement every draw is independent and the sizes multiply freely; without replacement each draw reduces the available pool by one, so the second factor decreases from 52 to 51.

Task 4 — Weekly Weather

Each day the weather can be in exactly one of three states: Sunny (S), Cloudy (C), or Rainy (R). Describe the sample space for one day, two days, and a full week (7 days), and give its size in each case.

One day

There are three possible states for a single day.

\Omega_1 = \{S,\, C,\, R\}, \quad |\Omega_1| = 3

Two days

Each outcome is an ordered pair of daily states, giving $3 \times 3 = 9$ equally likely combinations.

\Omega_2 = \{SS,\, SC,\, SR,\, CS,\, CC,\, CR,\, RS,\, RC,\, RR\}, \quad |\Omega_2| = 9

Seven days (one week)

Each outcome is an ordered 7-tuple of daily states. Applying the multiplication rule seven times:

|\Omega_7| = 3^7 = 2187

General formula

For $n$ days with $k$ weather states the sample space has $k^n$ elements.

|\Omega_n| = 3^n

Weather outcomes chart

Each extra day triples the number of sequences.

Weather outcomes = 3^k

Answer

|\Omega_1| = 3,\quad |\Omega_2| = 9,\quad |\Omega_7| = 3^7 = 2187

Explanation tip: The multiplication rule applies because each day's weather is chosen independently from the same set of 3 states; for 7 days the count is

3^7 = 2187

distinct weekly weather sequences.

Task 5 — Buffon's Needle (Sample Space)

A needle of length $L$ is dropped at random onto a floor ruled with parallel lines spaced $d$ apart (with $L \le d$ ). Describe the sample space and explain why it is continuous (uncountably infinite).

Identifying the random parameters

The position of the dropped needle is completely determined by two real-valued quantities: (1) the distance $x$ from the needle's center to the nearest line, and (2) the acute angle $\theta$ that the needle makes with the direction of the lines.

x \in \left[0,\, \frac{d}{2}\right], \quad \theta \in \left[0,\, \frac{\pi}{2}\right]

Sample space

An elementary outcome is the ordered pair $(x, \theta)$ . The sample space is the rectangle in the plane:

\Omega = \left\{(x,\theta) : x \in \left[0,\tfrac{d}{2}\right],\; \theta \in \left[0,\tfrac{\pi}{2}\right]\right\}

Why the space is continuous

Both $x$ and $\theta$ are real numbers ranging over intervals. Each interval contains uncountably many points (by the uncountability of $\mathbb{R}$ ), so $\Omega$ itself is uncountably infinite — it cannot be listed or enumerated.

|\Omega| = \mathfrak{c} \quad (\text{cardinality of the continuum})

Uniform probability measure

Because $\Omega$ is continuous, probabilities are assigned via area: the probability of any event $A \subseteq \Omega$ is proportional to the area of $A$ relative to the total area of $\Omega$ .

P(A) = \frac{\text{Area}(A)}{\text{Area}(\Omega)} = \frac{\text{Area}(A)}{\dfrac{d}{2} \cdot \dfrac{\pi}{2}} = \frac{4\,\text{Area}(A)}{\pi d}

Interactive simulation

Drop needles and watch the crossing rate. The continuous sample space $(x,\theta)$ has uncountably many outcomes.

Buffon's Needle — Monte Carlo

needles: 300L/d: 0.80

Crossings (red): 150 / 300

Empirical P: 0.500

Theoretical 2L/(πd): 0.509

π estimate = 2L·N/(d·C): 3.2000

Each needle crosses a line (red) when its half-projection onto the vertical exceeds the distance to the nearest line. The crossing rate converges to 2L/(πd).

Answer

\Omega = \left\{(x,\theta) : x \in \left[0,\tfrac{d}{2}\right],\; \theta \in \left[0,\tfrac{\pi}{2}\right]\right\}, \quad |\Omega| = \infty\text{ (uncountable)}

Explanation tip: Unlike discrete experiments, Buffon's needle has a continuous sample space because both the position

x

and angle

\theta

are real-valued; probabilities are computed as ratios of areas (a 2D Lebesgue measure), not ratios of counts.

Task 6 — Coin Probabilities

A fair coin is tossed. Compute the following probabilities: (a) heads on one toss; (b) exactly one head in two tosses; (c) exactly two heads in three tosses; (d) at least one tail in three tosses; (e) all tosses the same in three tosses.

a)(a) $A_1$ : heads on a single toss.
b)(b) $A_2$ : exactly one head in two tosses.
c)(c) $A_3$ : exactly two heads in three tosses.
d)(d) $B_3$ : at least one tail in three tosses.
e)(e) $C_3$ : all three tosses show the same face.

Part (a) — $A_1$: heads

The sample space has 2 equally likely outcomes; only 1 is heads.

A_1 = \{H\}, \quad P(A_1) = \frac{1}{2} = 0.5

Part (b) — $A_2$: exactly one head in two tosses

The sample space is $\{HH, HT, TH, TT\}$ with 4 equally likely outcomes. The favorable outcomes are $HT$ and $TH$ .

A_2 = \{HT,\, TH\}, \quad P(A_2) = \frac{2}{4} = \frac{1}{2} = 0.5

Part (c) — $A_3$: exactly two heads in three tosses

From $|\Omega_3| = 8$ outcomes, choose the ones with exactly 2 heads: $HHT, HTH, THH$ .

A_3 = \{HHT,\, HTH,\, THH\}, \quad P(A_3) = \frac{3}{8} = 0.375

Binomial check for (c)

Alternatively, using the binomial formula with $n=3$ , $k=2$ , $p=1/2$ :

P(A_3) = \binom{3}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^1 = 3 \cdot \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8}

Part (d) — $B_3$: at least one tail in three tosses

It is easier to use the complement: the only outcome with NO tail is $HHH$ .

P(B_3) = 1 - P(\{HHH\}) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875

Part (e) — $C_3$: all tosses the same

The favorable outcomes are $HHH$ (all heads) and $TTT$ (all tails), giving 2 out of 8.

C_3 = \{HHH,\, TTT\}, \quad P(C_3) = \frac{2}{8} = \frac{1}{4} = 0.25

Event probabilities chart

Comparing all five probabilities side by side.

Event probabilities

Answer

P(A_1)=\tfrac{1}{2},\; P(A_2)=\tfrac{1}{2},\; P(A_3)=\tfrac{3}{8},\; P(B_3)=\tfrac{7}{8},\; P(C_3)=\tfrac{1}{4}

Explanation tip: All five probabilities use the classical definition (favorable outcomes divided by total outcomes) on the finite equally-likely space; for

B_3

the complement rule is the most efficient approach.

Task 7 — Die Probabilities

A fair six-sided die is rolled once and twice. Compute the following probabilities: (a) an even number on one roll; (b) a number greater than 4 on one roll; (c) sum equals 7 on two rolls; (d) both dice show the same number; (e) sum is at least 10 on two rolls.

a)(a) $A_1$ : result is even on a single roll.
b)(b) $B_1$ : result is greater than 4 on a single roll.
c)(c) $A_2$ : sum of two dice equals 7.
d)(d) $B_2$ : both dice show the same number.
e)(e) $C_2$ : sum of two dice is at least 10.

Part (a) — $A_1$: even

Even faces are $\{2, 4, 6\}$ — three out of six equally likely outcomes.

A_1 = \{2,\,4,\,6\}, \quad P(A_1) = \frac{3}{6} = \frac{1}{2} = 0.5

Part (b) — $B_1$: greater than 4

Faces greater than 4 are $\{5, 6\}$ — two out of six.

B_1 = \{5,\,6\}, \quad P(B_1) = \frac{2}{6} = \frac{1}{3} \approx 0.333

Part (c) — $A_2$: sum equals 7

The pairs $(i,j)$ with $i+j=7$ are: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — exactly 6 favorable pairs out of 36.

P(A_2) = \frac{6}{36} = \frac{1}{6} \approx 0.167

Part (d) — $B_2$: doubles

Doubles are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — six pairs out of 36.

P(B_2) = \frac{6}{36} = \frac{1}{6} \approx 0.167

Part (e) — $C_2$: sum $\ge$ 10

Listing pairs with sum $\ge 10$ : $(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$ — six pairs out of 36.

P(C_2) = \frac{6}{36} = \frac{1}{6} \approx 0.167

Grid: sum = 7

The six favorable cells lie on the anti-diagonal.

	1	2	3	4	5	6
1	.	.	.	.	.	X
2	.	.	.	.	X	.
3	.	.	.	X	.	.
4	.	.	X	.	.	.
5	.	X	.	.	.	.
6	X	.	.	.	.	.

Die probabilities chart

All five event probabilities for one and two dice.

Answer

P(A_1)=\tfrac{1}{2},\; P(B_1)=\tfrac{1}{3},\; P(A_2)=\tfrac{1}{6},\; P(B_2)=\tfrac{1}{6},\; P(C_2)=\tfrac{1}{6}

Explanation tip: Parts (c), (d), and (e) each have exactly 6 favorable outcomes out of 36, so they all equal

1/6

— a useful coincidence to highlight when presenting to a professor to show mastery of enumeration.

Task 8 — Card Probabilities

One or two cards are drawn from a standard 52-card deck. Compute: (a) P(heart); (b) P(king); (c) P(not a face card); (d) P(both hearts) with replacement; (e) P(both hearts) without replacement.

a)(a) One card: probability it is a heart.
b)(b) One card: probability it is a king.
c)(c) One card: probability it is not a face card (Jack, Queen, King).
d)(d) Two cards with replacement: both are hearts.
e)(e) Two cards without replacement: both are hearts.

Part (a) — P(heart)

There are 13 hearts in a deck of 52 cards.

P(\text{heart}) = \frac{13}{52} = \frac{1}{4} = 0.25

Part (b) — P(king)

There are 4 kings (one per suit) in 52 cards.

P(\text{king}) = \frac{4}{52} = \frac{1}{13} \approx 0.077

Part (c) — P(not a face card)

Face cards are Jacks, Queens, Kings: $3 \times 4 = 12$ total. Non-face cards: $52 - 12 = 40$ .

P(\text{not face card}) = \frac{40}{52} = \frac{10}{13} \approx 0.769

Part (d) — both hearts, with replacement

With replacement, draws are independent; each has probability $1/4$ of being a heart.

P(\text{both hearts, with repl.}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625

Part (e) — both hearts, without replacement

The first draw: 13 hearts out of 52. Given the first was a heart, 12 hearts remain out of 51.

P(\text{both hearts, without repl.}) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17} \approx 0.059

Card probabilities chart

All five card event probabilities compared.

Answer

P(\heartsuit)=\tfrac{1}{4},\; P(K)=\tfrac{1}{13},\; P(\text{not face})=\tfrac{10}{13},\; P(\text{both}\heartsuit,\text{with})=\tfrac{1}{16},\; P(\text{both}\heartsuit,\text{without})=\tfrac{1}{17}

Explanation tip: Comparing parts (d) and (e) illustrates how removing a card from the deck makes subsequent draws dependent and slightly reduces the probability;

1/16 > 1/17

because replacement keeps all 13 hearts available for the second draw.

Task 9 — Weather Probabilities

Each day the weather is independently Sunny (S), Cloudy (C), or Rainy (R), each with probability $1/3$ . Compute: (a) P(both weekend days sunny); (b) P(Wed, Thu, Fri all rainy); (c) P(at least one sunny day in a week); (d) P(no rainy day in a week); (e) P(exactly two sunny days in a week).

a)(a) Both Saturday and Sunday are sunny.
b)(b) Wednesday, Thursday, and Friday are all rainy.
c)(c) At least one of the 7 days in the week is sunny.
d)(d) None of the 7 days in the week is rainy.
e)(e) Exactly two of the 7 days in the week are sunny.

Part (a) — both weekend days sunny

Saturday and Sunday are independent; each has probability $1/3$ of being sunny.

P(\text{Sat sunny} \cap \text{Sun sunny}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \approx 0.111

Part (b) — three consecutive rainy days

Wednesday, Thursday, and Friday are independent; each has probability $1/3$ of being rainy.

P(\text{Wed, Thu, Fri all rainy}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \approx 0.037

Part (c) — at least one sunny day in 7

Use the complement: "no sunny day in 7" means each day is Cloudy or Rainy, with probability $2/3$ each.

P(\text{at least one sunny}) = 1 - \left(\frac{2}{3}\right)^7 = 1 - \frac{128}{2187} = \frac{2059}{2187} \approx 0.941

Part (d) — no rainy day in 7

Each day must be Sunny or Cloudy (probability $2/3$ ). The 7 days are independent.

P(\text{no rainy day}) = \left(\frac{2}{3}\right)^7 = \frac{128}{2187} \approx 0.059

Part (e) — exactly two sunny days in 7

This is a binomial probability with $n=7$ , $k=2$ successes (sunny), $p=1/3$ , $q=2/3$ .

P(\text{exactly 2 sunny}) = \binom{7}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^5 = 21 \cdot \frac{1}{9} \cdot \frac{32}{243} = \frac{672}{2187} \approx 0.307

Weather probabilities chart

All five weather event probabilities compared.

Answer

\tfrac{1}{9},\; \tfrac{1}{27},\; \tfrac{2059}{2187},\; \tfrac{128}{2187},\; \tfrac{672}{2187}

Explanation tip: Parts (c) and (d) are complementary events summing to 1 (

2059/2187 + 128/2187 = 1

), which provides a useful self-check; part (e) applies the binomial model since each day is an independent trial with the same probability of success.

Task 10 — Buffon's Needle Probabilities

A needle of length $L \le d$ is dropped uniformly at random on a floor with parallel lines spaced $d$ apart. The needle's position is $(x, \theta)$ where $x \in [0, d/2]$ is the distance from the center to the nearest line and $\theta \in [0, \pi/2]$ is the acute angle with the lines. Compute: (a) P(needle crosses a line); (b) P(no crossing); (c) P(angle less than $\pi/6$ ); (d) P(center within $d/4$ of a line); (e) P(crosses AND angle $> \pi/4$ ).

a)(a) Event $A$ : the needle crosses a line.
b)(b) Event $B$ : the needle does not cross any line.
c)(c) Event $C$ : the angle satisfies $\theta < \pi/6$ .
d)(d) Event $D$ : the center is within $d/4$ of the nearest line.
e)(e) Event $E$ : the needle crosses a line AND the angle satisfies $\theta > \pi/4$ .

Area of the sample space

The sample space is a rectangle in the $(x, \theta)$ plane. Its total area is:

\text{Area}(\Omega) = \frac{d}{2} \cdot \frac{\pi}{2} = \frac{\pi d}{4}

Crossing condition

The needle crosses a line when the projection of its half-length onto the perpendicular to the lines exceeds $x$ , i.e., when $x \le \frac{L}{2}\sin\theta$ .

A = \left\{(x,\theta) : 0 \le x \le \frac{L}{2}\sin\theta,\; \theta \in \left[0,\frac{\pi}{2}\right]\right\}

Part (a) — P(crossing)

Compute the area of $A$ by integrating over $\theta$ , then divide by the total area.

P(A) = \frac{\displaystyle\int_0^{\pi/2} \frac{L}{2}\sin\theta\,d\theta}{\dfrac{\pi d}{4}} = \frac{\dfrac{L}{2}[-\cos\theta]_0^{\pi/2}}{\dfrac{\pi d}{4}} = \frac{\dfrac{L}{2}}{\dfrac{\pi d}{4}} = \frac{2L}{\pi d}

Part (b) — P(no crossing)

Event $B$ is the complement of $A$ .

P(B) = 1 - P(A) = 1 - \frac{2L}{\pi d}

Part (c) — P(angle $< \pi/6$)

The angle $\theta$ is uniform on $[0, \pi/2]$ , so the probability of any sub-interval is proportional to its length.

P(C) = \frac{\pi/6}{\pi/2} = \frac{1}{3} \approx 0.333

Part (d) — P(center within $d/4$ of a line)

The distance $x$ is uniform on $[0, d/2]$ , so $P(x \le d/4)$ is the fraction of the interval $[0,d/2]$ covered by $[0,d/4]$ .

P(D) = \frac{d/4}{d/2} = \frac{1}{2} = 0.5

Part (e) — P(crosses AND $\theta > \pi/4$)

Event $E$ restricts the crossing condition to $\theta \in (\pi/4, \pi/2)$ . Its area is computed by integrating the crossing bound over this sub-range:

\text{Area}(E) = \int_{\pi/4}^{\pi/2} \frac{L}{2}\sin\theta\,d\theta = \frac{L}{2}\left[-\cos\theta\right]_{\pi/4}^{\pi/2} = \frac{L}{2}\left(0 + \frac{\sqrt{2}}{2}\right) = \frac{L\sqrt{2}}{4}

Part (e) — final probability

Divide the favorable area by the total area of $\Omega$ :

P(E) = \frac{L\sqrt{2}/4}{\pi d/4} = \frac{L\sqrt{2}}{\pi d} = \frac{\sqrt{2}\,L}{\pi d}

Interactive: estimate the crossing probability

The empirical crossing rate converges to $2L/(\pi d)$ ; the slider changes $L/d$ .

Buffon's Needle — Monte Carlo

needles: 300L/d: 0.80

Crossings (red): 142 / 300

Empirical P: 0.473

Theoretical 2L/(πd): 0.509

π estimate = 2L·N/(d·C): 3.3803

Each needle crosses a line (red) when its half-projection onto the vertical exceeds the distance to the nearest line. The crossing rate converges to 2L/(πd).

Answer

P(A)=\dfrac{2L}{\pi d},\; P(B)=1-\dfrac{2L}{\pi d},\; P(C)=\dfrac{1}{3},\; P(D)=\dfrac{1}{2},\; P(E)=\dfrac{\sqrt{2}\,L}{\pi d}

Explanation tip: Buffon's needle is the canonical example of geometric probability — probabilities are areas in the parameter space

(x, \theta)

; the classical result

P(A) = 2L/(\pi d)

has historically been used to estimate

\pi

by repeated physical trials.

List 1 — Events & Probability

Task 1 — Coin Tossing

Task 2 — Rolling a Die

Task 3 — Drawing Cards

Task 4 — Weekly Weather

Task 5 — Buffon's Needle (Sample Space)

Task 6 — Coin Probabilities

Task 7 — Die Probabilities

Task 8 — Card Probabilities

Task 9 — Weather Probabilities

Task 10 — Buffon's Needle Probabilities