Computing Binomial Probabilities — Cumulative, Ranges and Shortcuts

For $X \sim B(n,p)$ with $q = 1-p$, the total probability splits across $r = 0,1,\dots,n$:

• Whole probability sums to 1: $\displaystyle\sum_{r=0}^{n} \binom{n}{r} p^r q^{\,n-r} = (p+q)^n = 1$.
• A compound event is a SUM of terms: e.g. $P(X \le 2) = P(0)+P(1)+P(2)$.
• Complement when it's shorter: $P(X \ge k) = 1 - P(X \le k-1)$. Pick whichever side has fewer terms.

The phrase-to-set dictionary: 'at least $k$' $= \{k, k+1, \dots, n\}$; 'at most $k$' $= \{0, 1, \dots, k\}$; 'more than $k$' $= \{k+1,\dots\}$; 'fewer/less than $k$' $= \{0,\dots,k-1\}$.

Turn the phrase into a sum, then compute each term with the PMF:

• At most one: $P(X \le 1) = P(0) + P(1) = q^n + n\,p\,q^{\,n-1}$.
• At least three (n=5): $P(X \ge 3) = P(3)+P(4)+P(5)$.
• 'Unable to solve less than two' (with p = solve): the candidate fails on 0 or 1 problem, i.e. SOLVES $n$ or $n-1$: $P = p^n + n\,p^{\,n-1}q$. Decide which event 'success' labels before you count.

Factor the common power to match the option form — e.g. $q^5 + 5pq^4 = q^4(q + 5p)$.

The complement collapses a whole tail to one term:

• At least one: $P(X \ge 1) = 1 - P(X = 0) = 1 - q^n$.
• Smallest n for a threshold: to force $P(X \ge 1) > t$, solve $1 - q^n > t \Rightarrow q^n < 1-t$, then take the least integer n. For a fair coin ($q=\tfrac12$) and $t = 0.99$: $\left(\tfrac12\right)^n < 0.01 \Rightarrow 2^n > 100 \Rightarrow n = 7$ (since $2^6=64,\ 2^7=128$).

This is exactly the trick behind 'probability of at least one defective bulb' $= 1 - (\text{good fraction})^n$.

Unpack the condition to an interval of integers, then choose the shorter side:

• $|X - a| \le b$ means $a - b \le X \le a + b$. For $X \sim B(6,\tfrac12)$, $|X-4|\le 2$ gives $2 \le X \le 6$.
• Since X can only be $0..6$, that block excludes just $X=0$ and $X=1$: $P(2 \le X \le 6) = 1 - P(0) - P(1)$.
• With $p=q=\tfrac12$, $P(r) = \binom{6}{r}/2^6$, so $P(0)+P(1) = \tfrac{1+6}{64} = \tfrac{7}{64}$ and the answer is $1 - \tfrac{7}{64} = \tfrac{57}{64}$.

Three recurring twists, each with its own hook:

• Even (or odd) number of successes: $\sum_{r\text{ even}}\binom{n}{r} = \dfrac{(1+1)^n + (1-1)^n}{2} = 2^{\,n-1}$. For a fair coin, $P(\text{even heads}) = \dfrac{2^{\,n-1}}{2^n} = \dfrac12$ (for any $n\ge 1$).
• Expected frequency over N repeats: if one experiment gives the event probability $P$, then repeating it $N$ times gives expected count $N \cdot P$ (this is the mean of a Binomial($N,P$)).
• Event fixed at a specific trial ('second success at the 3rd match'): FIX the last trial as a success, and require exactly the remaining successes among the earlier trials: $P = \binom{2}{1}p\,q \cdot p$ for a second win at match 3.
• Small-n sums like $P(X=1)+P(X=2)$ are just two PMF terms added directly.

Two steps: (1) count to get p, (2) apply the binomial to the required event.

• Count the favourable outcomes. For two-digit numbers 00–99 (100 in all) with digit product 24: $\{38, 46, 64, 83\}$, so $p = \tfrac{4}{100} = \tfrac{1}{25}$, $q = \tfrac{24}{25}$.
• Mind the sample space. '10–99' is 90 numbers, '00–99' is 100 — the denominator changes p, so read the range carefully.
• Then the binomial. With $n=4$, $P(X \ge 3) = \binom{4}{3}p^3q + p^4 = p^3(4q + p)$.