We can use the same binom.pmf() method from the scipy.stats library to calculate the probability of observing a range of values. As mentioned in a previous exercise, the binom.pmf method takes 3 values:

• x: the value of interest
• n: the sample size
• p: the probability of success

For example, we can calculate the probability of observing between 2 and 4 heads from 10 coin flips as follows:

import scipy.stats as stats

# calculating P(2-4 heads) = P(2 heads) + P(3 heads) + P(4 heads) for flipping a coin 10 times
print(stats.binom.pmf(2, n=10, p=.5) + stats.binom.pmf(3, n=10, p=.5) + stats.binom.pmf(4, n=10, p=.5))

Output:

# 0.366211

We can also calculate the probability of observing less than a certain value, let’s say 3 heads, by adding up the probabilities of the values below it:

import scipy.stats as stats

# calculating P(less than 3 heads) = P(0 heads) + P(1 head) + P(2 heads) for flipping a coin 10 times
print(stats.binom.pmf(0, n=10, p=.5) + stats.binom.pmf(1, n=10, p=.5) + stats.binom.pmf(2, n=10, p=.5))

Output:

# 0.0546875

Note that because our desired range is less than 3 heads, we do not include that value in the summation.

When there are many possible values of interest, this task of adding up probabilities can be difficult. If we want to know the probability of observing 8 or fewer heads from 10 coin flips, we need to add up the values from 0 to 8:

import scipy.stats as stats

stats.binom.pmf(0, n = 10, p = 0.5) + stats.binom.pmf(1, n = 10, p = 0.5) + stats.binom.pmf(2, n = 10, p = 0.5) + stats.binom.pmf(3, n = 10, p = 0.5) + stats.binom.pmf(4, n = 10, p = 0.5) + stats.binom.pmf(5, n = 10, p = 0.5) + stats.binom.pmf(6, n = 10, p = 0.5) + stats.binom.pmf(7, n = 10, p = 0.5) + stats.binom.pmf(8, n = 10, p = 0.5)

Output:

# 0.98926

This involves a lot of repetitive code. Instead, we can also use the fact that the sum of the probabilities for all possible values is equal to 1:

$\begin{aligned}
P(0\;to\;8\;heads) + P(9\;to\;10\;heads) = P(0\;to\;10\;heads) = 1 \\
P(0\;to\;8\;heads) = 1 - P(9\;to\;10\;heads)
\end{aligned}$

Now instead of summing up 9 values for the probabilities between 0 and 8 heads, we can do 1 minus the sum of two values and get the same result:

import scipy.stats as stats
# less than or equal to 8
1 - (stats.binom.pmf(9, n=10, p=.5) + stats.binom.pmf(10, n=10, p=.5))

Output:

# 0.98926