Choosing the Right Probability Formula

Choosing the Right Probability Formula — Moosa Academy

There is more than one probability formula, and each belongs to a different kind of situation. Most mistakes in probability are not arithmetic mistakes — they are choosing the wrong formula in the first place. Two questions are usually enough to decide.

Concept Two questions that decide everything

First: is the experiment run once, or repeated?
Second: if it is run once but several things are drawn, does the order matter?

Once, one outcome → simple probability

Once, several drawn together → combinations or permutations

Repeated, success or failure each time → the binomial formula

Example One toss: simple probability

A coin is tossed once. What is the probability of heads?

Step 1 — how many times? Once, with one outcome wanted

Step 2 — so this is simple probability:

P = \frac{\text{favourable}}{\text{possible}} = \frac{1}{2}

⟹

\tfrac{1}{2}

Nothing is repeated and nothing is drawn in a group, so no other formula is needed. Count the favourable outcomes and divide.

Example Repeated tosses: the binomial formula

A coin is tossed twice. What is the probability of heads both times?

Step 1 — how many times? Twice, and each toss is a success or a failure

Step 2 — so this is binomial, with

n = 2

k = 2

p = \tfrac{1}{2}

P = \binom{n}{k} p^{k} (1-p)^{n-k}

Step 3 — substitute:

\binom{2}{2} \left(\tfrac{1}{2}\right)^{2} \left(\tfrac{1}{2}\right)^{0} = 1 \times \tfrac{1}{4} \times 1

⟹

\tfrac{1}{4}

The word "twice" is what forces the change of formula. Repetition with a success-or-failure result each time is exactly what the binomial formula was built for.

Example Drawing a group: combinations

There are 20 secondary students and 20 middle students. Six students are drawn. What is the probability that three come from each group?

Step 1 — how many times? Once, but six students are drawn together

Step 2 — does the order matter? No, a group of six is the same group whatever order it is listed in — so use combinations

Step 3 — favourable:

\binom{20}{3} \times \binom{20}{3} = 1140 \times 1140 = 1{,}299{,}600

Step 4 — possible:

\binom{40}{6} = 3{,}838{,}380

Step 5 — divide:

\dfrac{1{,}299{,}600}{3{,}838{,}380} \approx 0.339

⟹ about 33.9%

Had the order mattered — for instance if the six students were being placed in ranked positions — permutations $P$ would replace combinations $C$ .

Note The three formulas together

Simple — one experiment, one outcome: favourable ÷ possible. Example: one roll of a die.

Binomial — repeated trials, success or failure each time. Example: five coin tosses.

Combinations or permutations — one experiment, several items drawn together. Example: choosing students from groups.

Reading the question for these signals costs a few seconds and saves the whole solution. Decide the type first, then apply the formula.

Summary

Ask first how many times the experiment is run.
Once with one outcome wanted → simple probability, favourable ÷ possible.
Repeated with success or failure each time → the binomial formula.
Once but several items drawn together → combinations, or permutations if order matters.
Two heads in two tosses is binomial: 1/4.
Three from each group of twenty, six drawn, is C(20,3)² ÷ C(40,6) ≈ 33.9%.