Elementary Statistics and Probability

Elementary statistics summarises a list of numbers by a centre (mean, median, mode) and a spread (range, variance, standard deviation). Elementary probability measures how likely an event is, usually as a ratio of counts or as a product of the probabilities of successive steps. The calculations that follow are of this kind: an average that changes when an item is added or replaced, the standard deviation of a small list, the effect of adding a constant to every value, or the chance of drawing two red balls.

Watch for the median of an even-sized list, population versus sample standard deviation, sampling with versus without replacement, and "at least one", which is usually easiest through the complement.

Variance convention. These notes use the population formula (divide by $n$) unless a question says "sample" or "unbiased", in which case divide by $n - 1$. A useful identity is $\sigma^2 = \overline{x^2} - \bar{x}^2$ (mean of squares minus square of mean).

Linear transformation. If every value becomes $y_i=ax_i+b$, then $\bar y=a\bar x+b$, the median becomes $a\operatorname{median}(x)+b$, and the standard deviation becomes $|a|\sigma$. For $a\ne0$, the mode values transform one-to-one by the same rule. For $a=0$, all observations equal $b$, so $b$ is the unique mode. Adding a constant shifts the centre and leaves the spread unchanged.

Mutually exclusive is different from independent. Two events with positive probability that cannot happen together are dependent: knowing one happened makes the other impossible.

Totals make averages easy. If $n$ values have mean $m$, their total is $nm$. Adding, removing or replacing values changes the total by a known amount, and the new mean is the new total over the new count. Combining groups of sizes $n_1, n_2$ with means $m_1, m_2$ gives the weighted mean $\frac{n_1 m_1 + n_2 m_2}{n_1 + n_2}$.

Why $n - 1$. Deviations from the sample mean sum to zero, so only $n - 1$ of them are free; dividing by $n - 1$ makes the sample variance an unbiased estimate of the population variance. For the arithmetic of a GA question, the only point is which divisor the question wants.

Complement for "at least one". For independent trials each succeeding with probability $p$, $P(\text{no success in } n) = (1 - p)^n$, so $P(\text{at least one}) = 1 - (1 - p)^n$.

Binomial count. The probability of exactly $k$ successes in $n$ independent trials is $\binom{n}{k} p^k (1 - p)^{n-k}$: choose which $k$ trials succeed, then multiply.

Example 1 (centre and spread). Data: 4, 8, 6, 5, 3, 8, 9. Sorted: 3, 4, 5, 6, 8, 8, 9.

Example 2 (median of an even-sized list). For 3, 7, 8, 12, the median is $(7 + 8)/2 = 7.5$.

Example 3 (combined mean). Class A has 30 students with mean 60; class B has 20 with mean 70. Combined mean: $\frac{30 \times 60 + 20 \times 70}{50} = \frac{3200}{50} = 64$.

Example 4 (replacing a value). Ten numbers have mean 25, so their total is 250. One number, 40, is replaced by 10. New total $250 - 40 + 10 = 220$; new mean $22$.

Example 5 (standard deviation). Data: 2, 4, 4, 4, 5, 5, 7, 9 ($n = 8$).

Example 6 (linear transformation). Each value of Example 5 is multiplied by 3 and then increased by 2. New mean $3 \times 5 + 2 = 17$; new population standard deviation $3 \times 2 = 6$. The added 2 does not change the spread.

Example 7 (two dice). Two fair dice are rolled (36 equally likely ordered outcomes).

Example 8 (addition rule). One card from a standard 52-card deck. $P(\text{king or heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$; the king of hearts is subtracted once because it was counted twice.

Example 9 (with and without replacement). A bag has 5 red and 3 blue balls; two are drawn.

Example 10 (at least one). Each of 4 independent shots hits with probability 0.3. $P(\text{at least one hit}) = 1 - 0.7^4 = 1 - 0.2401 = 0.7599$.

Example 11 (conditional). A family has two children, each equally likely to be a boy or a girl, independently. Given that at least one is a boy, the sample space is $\{BB, BG, GB\}$, so $P(\text{both boys}) = 1/3$, not $1/2$.

Example 12 (binomial). Exactly 2 heads in 4 fair tosses: $\binom{4}{2}/2^4 = 6/16 = 3/8$.

Example 13 (expected value). A game pays 10 with probability $1/6$ and costs 1 otherwise. $E = 10 \times \frac{1}{6} - 1 \times \frac{5}{6} = \frac{5}{6} \approx 0.83$ per play.

Example 14 (frequency table). Marks 1 to 5 were scored by 25 students with frequencies 3, 5, 8, 6, 3.

Example 15 (grouped data). 40 values fall in the classes 0 to 10, 10 to 20, 20 to 30, 30 to 40, 40 to 50 with frequencies 5, 8, 12, 10, 5 (each class includes its lower end).

Example 16 (two-stage experiment and reversed conditional). Urn I has 3 red and 2 blue balls; urn II has 1 red and 3 blue. A fair coin chooses the urn, then one ball is drawn.

Example 17 (table and direction of conditioning). Of 100 students, 60 took science and 45 of them passed; 40 took arts and 20 of them passed. $P(\text{passed} \mid \text{science}) = \frac{45}{60} = \frac{3}{4}$, but $P(\text{science} \mid \text{passed}) = \frac{45}{65} = \frac{9}{13}$. Identify the condition (the group you restrict to) before dividing.

Example 18 (several draws without replacement). Three balls are drawn without replacement from 5 red and 3 blue. $P(\text{at least one red}) = 1 - \frac{\binom{3}{3}}{\binom{8}{3}} = 1 - \frac{1}{56} = \frac{55}{56}$. With replacement the answer would be $1 - \left(\frac{3}{8}\right)^3 = \frac{485}{512}$, slightly smaller.