4.10 Probability for Quantitative Aptitude

Definition and Range

For an experiment whose outcomes are equally likely,

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

with $0 \leq P(E) \leq 1$. An impossible event has probability 0 and a certain event probability 1. The complement satisfies

$P(\bar{E}) = 1 - P(E)$

This section covers the aptitude form of probability set on Paper-I. The engineering treatment — random variables, and the binomial, Poisson and normal distributions — belongs to the Engineering Mathematics chapter of Paper-II.

The Standard Sample Spaces

Commit these to memory; almost every question uses one of them.

Coins

Tossing $n$ coins gives $2^{n}$ outcomes. For 2 coins: HH, HT, TH, TT. For 3 coins: 8 outcomes.

Dice

One die has 6 outcomes; two dice have 36. The distribution of the sum is worth tabulating:

A sum of 7 is the most likely, with probability $6/36 = 1/6$. Doubles occur in 6 of 36 ways, again $1/6$.

A standard deck of cards

The most-missed figure is that there are 12 face cards, not 16 — the ace is not a face card in the usual convention.

The Addition Rule

$P(A \cup B) = P(A) + P(B) - P(A\cap B)$

Example. Drawing one card, find the probability that it is a king or a heart.

$P = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

The subtraction removes the king of hearts, counted in both terms.

The Multiplication Rule: Replacement Matters

For independent events, $P(A\cap B) = P(A)P(B)$. Successive draws are independent only with replacement.

Example. A bag holds 5 red and 3 blue balls. Two are drawn.

With replacement, the composition is restored each time:

$P(\text{both red}) = \frac{5}{8}\times\frac{5}{8} = \frac{25}{64}$

Without replacement, the second draw is conditional on the first:

$P(\text{both red}) = \frac{5}{8}\times\frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Both the numerator and the denominator fall by one. Reading whether the question says with replacement is the single most important step.

The combination method

Without replacement, an equivalent and often faster route uses combinations:

$P(\text{both red}) = \frac{^{5}C_{2}}{^{8}C_{2}} = \frac{10}{28} = \frac{5}{14}$

This matches, and it generalises cleanly. For 3 balls drawn from the same bag, the probability that exactly 2 are red is

$\frac{^{5}C_{2}\times{}^{3}C_{1}}{^{8}C_{3}} = \frac{10\times3}{56} = \frac{30}{56} = \frac{15}{28}$

The Complement Shortcut

Whenever a question contains the phrase at least one, compute the complement.

$P(\text{at least one}) = 1 - P(\text{none})$

Example. Three coins are tossed. Find the probability of at least one head.

$P = 1 - P(\text{no head}) = 1 - \frac{1}{8} = \frac{7}{8}$

Enumerating the seven favourable outcomes works but takes several times as long.

Example. Two dice are thrown. Find the probability of at least one six.

$P = 1 - \left(\frac{5}{6}\right)^{2} = 1 - \frac{25}{36} = \frac{11}{36}$

Note that $11/36$, not $12/36$, is correct — adding $1/6 + 1/6$ double-counts the double six.

Odds

Odds in favour of an event are the ratio of favourable to unfavourable outcomes; odds against reverse the ratio.

$\text{If odds in favour are } a:b, \text{ then } P = \frac{a}{a+b}$

Example. If the odds against an event are 4 : 3, then $P = 3/7$.

Worked Example Set

The leap-year item links directly to the odd-days method in the Clocks and Calendars section — a non-leap year gives $1/7$ because it carries only one odd day.

Exam Discipline

1. Write the total number of outcomes first.
2. Check for the words with replacement and at least.
3. Prefer the complement when the phrase at least one appears.
4. Reduce the fraction; options are always in lowest terms.
5. Sanity check that the answer lies between 0 and 1 — a result outside that range signals an arithmetic slip, and with no negative marking there is no reason not to correct and answer.