The binomial distribution

Ten trials, probability 0.3 each time. Choose an event and the bars it covers light up. Most lost marks on this sub-topic are a shaded bar too many or too few.

k = 3

0.2668probability

3mean np

2.1variance np(1−p)

Read the shading, not the words. The words are what trip people up.

When is it binomial?

A fixed number of trials, n.
Each trial has two outcomes, success or failure.
The trials are independent.
The probability p is the same every time.

Drawing counters without replacement is not binomial, because p changes after each draw. That is the most common situation students force into a binomial, and it is exactly the tree diagram case from 4.6.

Mean and variance

mean= np = 10 × 0.3 = 3 the same n × P as 4.5 variance= np(1 − p) = 10 × 0.3 × 0.7 = 2.1 standard deviation= √2.1 ≈ 1.449 the square root, not the variance itself

Reading the phrasing

The question says	In symbols	Bars covered
exactly 3	P(X = 3)	just 3
at most 3, no more than 3	P(X ≤ 3)	0, 1, 2, 3
fewer than 3, less than 3	P(X ≤ 2)	0, 1, 2
at least 3, 3 or more	P(X ≥ 3)	3 up to 10
more than 3	P(X ≥ 4)	4 up to 10

"At least one" is always 1 − P(X = 0). Shading nine bars and adding them is nine chances to slip; the complement is one calculation. For B(10, 0.3), P(X ≥ 1) = 1 − 0.0282 = 0.9718.

Worked example

P(X = 3)= ¹⁰C₃(0.3)³(0.7)⁷ = 120 × 0.027 × 0.0824 ≈ 0.2668 P(X ≤ 2)= 0.3828 three bars added, best done on the calculator P(X ≥ 4)= 1 − P(X ≤ 3) = 0.3504 note the 3, not the 4

Your turn

1. For B(10, 0.3), find P(X = 3) to 4 decimal places.

2. Still B(10, 0.3). Find P(X ≥ 4) to 4 decimal places.

3. Which of these is not binomial?

Rolling a die 20 times and counting sixes. Drawing 3 counters from a bag without replacement and counting reds. Asking 50 people a yes or no question, each answering independently with the same probability. Flipping a coin 12 times and counting heads.

Where the marks go

Writing X ~ B(n, p) with the numbers in it. It states the model, and it is often an explicit mark before any calculation.

Translating the words into the right inequality. "More than 3" is X ≥ 4, and getting that boundary wrong loses the whole answer even with perfect technology use.

Using the complement for "at least one". Expected, and much safer than adding bars.

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