13.2 Calculating binomial probabilities — Learn

Prerequisites: the binomial formula and the four conditions from Section 13.1; the complement rule from Section 12.1.

By the end of this section you can: find $P(X = r)$ for a binomial variable; find an at-most probability $P(X \le r)$ ; find an at-least probability $P(X \ge r)$ using the complement; and find a between probability $P(a \le X \le b)$ .

1. From one value to a range

The formula from Section 13.1 gives the probability of exactly one value of $X$ :

P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}.

Most questions ask for a range of values. The probabilities of the individual values are added, because the values do not overlap. For a binomial variable,

P(X \le r) = P(X = 0) + P(X = 1) + \dots + P(X = r).

Example 1. The random variable $X \sim B(10, 0.4)$ . Find $P(X \le 2)$ .

The possible values 0, 1 and 2 are the ones we need. Work out each term.

P(X = 0) = (0.6)^{10} = 0.0060466 \ (4 \text{ d.p.}),

P(X = 1) = \binom{10}{1}(0.4)(0.6)^9 = 10 \times 0.4 \times 0.0100777 = 0.0403108 \ (4 \text{ d.p.}),

P(X = 2) = \binom{10}{2}(0.4)^2(0.6)^8 = 45 \times 0.16 \times 0.0167962 = 0.1209324 \ (4 \text{ d.p.}).

Add the three terms:

P(X \le 2) = 0.0060466 + 0.0403108 + 0.1209324 = 0.1673 \ (4 \text{ d.p.}).

2. At least, using the complement

For an at-least probability, the values run to the top of the range. There are two routes.

Add the terms from $r$ up to $n$ .
Use the complement: $P(X \ge r) = 1 - P(X \le r - 1)$ .

Choose the route with fewer terms. The complement rule is

P(X \ge r) = 1 - P(X < r) = 1 - P(X \le r - 1).

The subtraction uses $r - 1$ , not $r$ . The event $X \ge r$ leaves out every value up to and including $r - 1$ .

Example 2. The random variable $X \sim B(10, 0.4)$ . Find $P(X \ge 8)$ .

Adding the terms from 8 to 10 uses three terms:

P(X = 8) = \binom{10}{8}(0.4)^8(0.6)^2 = 45 \times 0.0006554 \times 0.36 = 0.0106168,

P(X = 9) = \binom{10}{9}(0.4)^9(0.6) = 10 \times 0.0002621 \times 0.6 = 0.0015729,

P(X = 10) = (0.4)^{10} = 0.0001049.

Add:

P(X \ge 8) = 0.0106168 + 0.0015729 + 0.0001049 = 0.0123 \ (4 \text{ d.p.}).

The complement route would need the seven terms from 0 to 7, so adding the three terms is shorter.

3. Between two values

For a between probability, add the terms from the lower value to the upper value:

P(a \le X \le b) = P(X \le b) - P(X \le a - 1).

Example 3. The random variable $X \sim B(12, 0.3)$ . Find $P(3 \le X \le 5)$ .

The values are 3, 4 and 5. Work out the three terms.

P(X = 3) = \binom{12}{3}(0.3)^3(0.7)^9 = 220 \times 0.027 \times 0.0403536 = 0.2397 \ (4 \text{ d.p.}),

P(X = 4) = \binom{12}{4}(0.3)^4(0.7)^8 = 495 \times 0.0081 \times 0.0576480 = 0.2311 \ (4 \text{ d.p.}),

P(X = 5) = \binom{12}{5}(0.3)^5(0.7)^7 = 792 \times 0.00243 \times 0.0823543 = 0.1585 \ (4 \text{ d.p.}).

Add:

P(3 \le X \le 5) = 0.2397 + 0.2311 + 0.1585 = 0.6293 \ (4 \text{ d.p.}).

4. Progress check

Return to the chapter problem. The supplier claims that each of the 20 seeds germinates with probability 0.8, and the seeds germinate independently. Let $X$ be the number that germinate. Then $X \sim B(20, 0.8)$ .

Find the probability that 13 or fewer seeds germinate. Use the complement and the terms from 14 to 20, or a calculator.

Answer

P(X \le 13) = 1 - P(X \ge 14) = 1 - \left(P(X = 14) + \dots + P(X = 20)\right) \approx 0.0867.

So 13 or fewer seeds germinate in about 8.7% of packets of this kind. Keep this value for the Review.

5. Try it: shading a range

Predict first. The figure shows the bars of a binomial distribution for $X \sim B(n, p)$ , with a value $r$ you can change. Before you move anything, predict how the shaded area changes when you switch from "exactly $r$ " to "at most $r$ " to "at least $r$ " for the same $n$ , $p$ and $r$ .

Test. Set $n = 10$ and $p = 0.4$ . Choose "at most" and set $r = 2$ . Read the probability, then change to "exactly" with the same $r$ . Compare the two readings. Now use "at least" with $r = 8$ and compare it with $1 - P(X \le 7)$ .

Explain. Write two sentences. The first explains why "at most" and "at least" together cover the whole range but overlap at $r$ . The second explains why $P(X \ge r) = 1 - P(X \le r - 1)$ uses $r - 1$ .

Watch out. Three mistakes are common.

Subtracting the wrong value. For $P(X \ge r)$ , subtract $P(X \le r-1)$ , not $P(X \le r)$ . Using $P(X \le r)$ removes the value $r$ as well, so the answer is too small.

Leaving out a term. For $P(3 \le X \le 5)$ , the values are 3, 4 and 5. Include both ends. The answer is not $P(X < 5)$ .

Rounding at every step. Keep the individual terms to four decimal places or more, and round once at the end. Rounding each term to two places can move the final answer by more than the accuracy you want.

Re-check: say the inequality in words before you calculate. "At least 8" means 8, 9, 10, and so on. "Between 3 and 5" means 3, 4 and 5.

You should now be able to

add individual binomial terms to find $P(X \le r)$ ;
use $P(X \ge r) = 1 - P(X \le r-1)$ and choose the shorter calculation;
find $P(a \le X \le b)$ by adding the terms from $a$ to $b$ ;
keep enough accuracy and round once.