Simple Probability or the Binomial Formula

Simple Probability or the Binomial Formula — Moosa Academy

When an experiment is repeated, there are two quite different questions you might be asked: do all of the trials succeed, or does some particular number of them succeed? The first is a simple multiplication. The second needs the binomial formula, and the reason is worth understanding rather than memorising.

Concept The question that separates them

One trial only → simple probability: favourable ÷ possible.

Several trials, all must succeed → multiply the probabilities directly.

Several trials, a set number must succeed → the binomial formula.

The dividing question is simply: all of them, or a certain number of them? Everything in this lesson follows from that one distinction.

Example One question: simple probability

A multiple-choice question has four options. Guessing, what is the probability of answering correctly?

Step 1 — one trial only

Step 2 — favourable is 1, possible is 4

P = \frac{1}{4}

⟹

\tfrac{1}{4}

, or 25%

Example Two questions, both right: multiply

Two such questions are answered by guessing. What is the probability that both are right?

Step 1 — every trial must succeed, so there is only one way for this to happen

Step 2 — multiply the probabilities:

\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

⟹

\tfrac{1}{16}

, or 6.25%

No extra factor is needed here. "Both right" describes exactly one pattern — right then right — so multiplying is the whole calculation.

Example Three questions, exactly two right: binomial

Three questions are answered by guessing. What is the probability that exactly two are right?

Multiplying alone would count only one of these. There are three different patterns that give exactly two correct, and each is equally likely — so the answer must be three times larger.

Step 1 — the values:

n = 3

x = 2

p = \tfrac{1}{4}

q = \tfrac{3}{4}

Step 2 — the formula:

P = \binom{n}{x} p^{x} q^{\,n-x}

Step 3 — the count of patterns:

\binom{3}{2} = 3

Step 4 — substitute:

3 \times \left(\tfrac{1}{4}\right)^{2} \times \tfrac{3}{4} = 3 \times \tfrac{1}{16} \times \tfrac{3}{4}

Step 5 — multiply out:

\tfrac{9}{64} \approx 0.141

⟹

\tfrac{9}{64}

, or about 14.1%

The binomial coefficient $\binom{3}{2}$ is not decoration — it is exactly the count of the patterns drawn above. That is the whole difference between this case and the previous one.

Note Why "all of them" needs no coefficient

The binomial formula covers the earlier case too. For two questions both right, $n = 2$ and $x = 2$ :

\binom{2}{2} = 1

, so the coefficient does nothing

1 \times \left(\tfrac{1}{4}\right)^{2} \times \left(\tfrac{3}{4}\right)^{0} = \tfrac{1}{16}

⟹ the same answer as before

When every trial must succeed there is only one pattern, so the coefficient is 1 and the formula collapses back into plain multiplication. Asking for some of the trials is what makes the coefficient earn its place.

Summary

One trial → simple probability: one question of four options gives 1/4.
All trials must succeed → multiply: two both right gives 1/16.
A set number must succeed → binomial: P = C(n,x) pˣ q⁽ⁿ⁻ˣ⁾.
Exactly two right out of three gives 3 × (1/4)² × (3/4) = 9/64 ≈ 14.1%.
The coefficient counts the patterns — here the three ways two answers can be the right ones.
When all must succeed the coefficient is 1, so the formula becomes ordinary multiplication.