Probability Fundamentals & Geometric Probability

Two fair dice are rolled. What is the sample space, and how likely is the sum $7$?

Solution. List ordered pairs $(a,b)$ so each is equally likely: $|S| = 6 \cdot 6 = 36$. The sum is $7$ for $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so $|E| = 6$ and

If instead we had used the $11$ possible sums $2,3,\dots,12$ as our sample space, those outcomes are not equally likely, and the counting formula would give a wrong answer.

One card is drawn from a standard $52$-card deck. Find $P(\text{a face card or a heart})$.

Solution. Face cards: $12$. Hearts: $13$. Overlap (face-card hearts): $3$. By inclusion–exclusion the favorable count is $12 + 13 - 3 = 22$, so

A jar has $4$ red and $6$ blue marbles. One marble is drawn. Find $P(\text{red})$.

Solution. $P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5}$. Contest answers are almost always expected as reduced fractions.

A fair coin is flipped $5$ times. Find $P(\text{at least one head})$.

Solution. The complement is “no heads,” i.e. all $5$ tails, with probability $\left(\tfrac12\right)^5 = \tfrac{1}{32}$. Hence

Counting the $5$ separate “exactly $k$ heads” cases would be far more work.

Three people each pick a day of the week at random. Find $P(\text{at least two pick the same day})$.

Solution. Complement: all three days distinct. Total $= 7^3 = 343$; all-distinct $= 7\cdot 6\cdot 5 = 210$. So

A club of $6$ boys and $4$ girls chooses a $3$-person committee at random. Find $P(\text{exactly } 2 \text{ girls})$.

Solution. Total committees: $\binom{10}{3}=120$. Favorable: choose $2$ of $4$ girls and $1$ of $6$ boys, $\binom{4}{2}\binom{6}{1}=6\cdot 6=36$. So

Five cards are dealt from a standard deck. Find $P(\text{all five are hearts})$.

Solution. Total $5$-card hands: $\binom{52}{5}$. All-heart hands: $\binom{13}{5}$. So

Order is irrelevant to a hand, so combinations appear top and bottom.

A bag has $3$ red and $5$ green marbles. Two are drawn with replacement. Find $P(\text{both red})$.

Solution. Replacement restores the bag, so the draws are independent, each with $P(\text{red})=\tfrac{3}{8}$:

Same bag ($3$ red, $5$ green), but the two marbles are drawn without replacement. Find $P(\text{both red})$.

Solution. The second draw depends on the first:

Fewer red marbles and fewer total marbles remain, so the second factor shrinks.

A dart lands at a uniformly random point in a $4\times 4$ square. Find the probability it lands inside the inscribed circle of radius $2$ (figure above).

Solution. Favorable area $=\pi(2)^2 = 4\pi$; total area $=16$. So

Geometric probabilities need not be rational — they inherit whatever constants ($\pi$, radicals) the geometry contains.

A point $X$ is chosen uniformly on the segment from $0$ to $12$. Find $P(X > 9)$.

Solution. Favorable length $= 12 - 9 = 3$; total length $= 12$. So $P = \dfrac{3}{12} = \dfrac{1}{4}.$

Two friends each arrive at the café at a uniformly random time between 12:00 and 1:00, independently, and each waits $15$ minutes. Find the probability that they meet.

Solution. Let $x,y \in [0,60]$ be their arrival minutes. The sample space is the $60\times 60$ square, area $3600$. They meet exactly when $|x-y|\le 15$, the band between the lines $y = x+15$ and $y = x-15$. The two failure corners are right triangles with legs $60-15 = 45$, so the failure area is $2\cdot\tfrac12\cdot 45^2 = 45^2 = 2025$. Thus

Two independent random times $\Rightarrow$ a point in a square $\Rightarrow$ area ratio. This “two random reals” setup is the signature of area-based geometric probability.

A stick of length $1$ is broken at a uniformly random point. What is the probability the longer piece is at least twice the shorter?

Solution. Let the break be at $X\in[0,1]$. The longer piece is at least twice the shorter exactly when the shorter piece is at most $\tfrac13$, i.e. $X\le\tfrac13$ or $X\ge\tfrac23$. Favorable length $=\tfrac13+\tfrac13=\tfrac23$, so $P=\dfrac{2}{3}$. One random point $\Rightarrow$ length ratio.

A standard $52$-card deck is shuffled uniformly at random. Find the probability that the ace of spades appears somewhere above the king of hearts, and that both appear above the two of clubs.

Solution. Ignore all $49$ other cards: by symmetry the three named cards appear among themselves in each of the $3!=6$ relative orders equally likely. Exactly one of those orders is “ace of spades, then king of hearts, then two of clubs.” Hence

The $49$ irrelevant cards never enter the computation — symmetry collapses the whole shuffle to the relative order of the three cards we care about.

From a bag of chips numbered $1$ through $10$, two chips are drawn at random. Find $P(\text{the two numbers differ by exactly }1)$, and then the same probability for chips $1$ through $n$.

Solution. Total unordered pairs: $\binom{10}{2}=45$, all equally likely. The consecutive pairs are $\{1,2\},\{2,3\},\dots,\{9,10\}$: $9$ of them, one for each possible smaller number $k = 1, \ldots, 9$. So $P=\dfrac{9}{45}=\dfrac{1}{5}$. In general the correspondence $\{k,k+1\}\leftrightarrow k$ matches the consecutive pairs with $\{1,\ldots,n-1\}$, so there are $n-1$ of them and $P = \dfrac{n-1}{\binom{n}{2}}=\dfrac{2}{n}$.

A point $(x,y)$ is chosen uniformly in the unit square $[0,1]^2$.

(a) Find $P\!\left(x+y \le \tfrac12\right)$.   (b) Find the expected value of $x+y$.

Solution. (a) The whole square has area $1$. The favorable set $x+y\le\tfrac12$ is a right triangle with legs $\tfrac12$, area $\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$. A clean shape, so use a ratio:

(b) An average, so use symmetry and linearity of expectation (previewed below). The mirror $x \mapsto 1-x$ sends the uniform point to another uniform point, so $\mathbb{E}[x] = \mathbb{E}[1-x] = 1-\mathbb{E}[x]$, forcing $\mathbb{E}[x] = \tfrac12$; the same holds for $y$. Adding the two averages,

Part (a) wanted the size of a region (ratio); part (b) wanted an average (symmetry plus linearity).

Real numbers $x,y,z$ are chosen independently and uniformly from $[0,1]$. Find the probability that they can be the side lengths of the three edges meeting at a corner of a box whose space diagonal is at most $1$, i.e. $x^2+y^2+z^2 \le 1$.

Solution. The sample space is the unit cube, volume $1$. The favorable region is the part of the cube inside the sphere of radius $1$ centered at the origin — exactly one octant of that ball:

Three independent uniform reals put us in a cube (volume ratio); the curved boundary is a sphere, so we lean on the known ball-volume formula — no heavier machinery needed.

Four letters are placed at random into four pre-addressed envelopes, one per envelope. Find the probability that at least one letter lands in its correct envelope.

Solution. Let $A_i$ be the event that letter $i$ is correct. By symmetry $P(A_i)=\tfrac{3!}{4!}=\tfrac14$; each pairwise $P(A_i\cap A_j)=\tfrac{2!}{4!}=\tfrac{1}{12}$; each triple $\tfrac{1!}{4!}=\tfrac{1}{24}$; the quadruple $\tfrac{1}{24}$. Inclusion–exclusion:

Equivalently, $1$ minus the derangement probability $\tfrac{9}{24}=\tfrac{3}{8}$. As $n\to\infty$ this “at least one fixed point” probability tends to $1-e^{-1}$.

Each of $10$ independent components fails on a given day with probability $0.002$. Bound the probability that the system, which fails if any component fails, fails today.

Solution. Exact inclusion–exclusion is messy, but the union bound is instant:

The true value $1-(0.998)^{10}\approx 0.0198$ is just under the bound — when the events are rare and barely overlap, the union bound is both quick and tight.

A biased coin shows heads with probability $p=\tfrac13$. It is flipped repeatedly and independently.

(a) In $5$ flips, find $P(\text{exactly }2\text{ heads})$.   (b) Find $P(\text{the first head is on flip }4)$.

Solution. (a) Binomial with $n=5,\,k=2$:

(b) Geometric with $p=\tfrac13$: three failures then a success,

Identify the family (fixed number of trials $\Rightarrow$ binomial; wait for the first success $\Rightarrow$ geometric) and the formula writes itself.

A fair die is rolled once. Find the expected value of the number shown. Then find the expected number of heads in $5$ flips of the coin with $p=\tfrac13$.

Solution. Definition. The expected value of a discrete random variable is the probability-weighted average of its values, $\mathbb{E}[X]=\sum_x x\,P(X=x)$. For the die (discrete uniform on $1$–$6$):

For the coin, rather than sum the binomial terms, use linearity of expectation: each flip contributes an expected $p=\tfrac13$ head, so

In general $\mathbb{E}[B(n,p)]=np$ and $\mathbb{E}[\text{geometric}(p)]=\tfrac1p$. Unit 9 develops linearity of expectation into a power tool; for now, note it lets you average without ever computing the full distribution.