We suggest you read through the lesson below to better understand the formula, but if you just want a quick example or 2 of the formula, here you go :

If there are 4 possible outcomes, and you want exactly 1 of them to occur, the formula is

$
_4 C _1 \cdot p(\text{ success})^1 \cdot p(\text{fail})^{(4-1)}
\\
_4 C _1 \cdot p(\text{ success})^1 \cdot p(\text{fail})^3
$

Consider the spinner below.

Spinner

If we spin 7 times, what is the probability we will get an even number exactly exactly 2 times?

$
_{\text{total trials}} C _{\text{exactly n}} \cdot p(success) ^{\text{success times}} \cdot p (fail)^{(\text{fail times})}
\\
$

$
_7 C _2 \cdot p \left( \frac{1}{3} \right)^2 \cdot p \left( \frac{2}{3} \right)^{(7-2)}
\\
_7 C _2 \cdot p \left( \frac{1}{3} \right)^2 \cdot p \left( \frac{2}{3} \right)^{5}
$

In the next part, we will explain the concepts behind this formula. If you'd like to just practice applying the formula, just go to our practice problems at the bottom of this page.

Part I. How we get the formula for multiple trial Probability

The formula to calculate the probability that an event will occur exactly n times over multiple trials is intricately tied to the formula for combinations. This may be a surprise at first, but upon examination there is a clear connection between combinations and multiple trial probabilities.

So, if you flip a coin, you have a $$\frac 1 2 $$ probability of getting heads.

What if we flip the coin twice?

What is the probablilty that we get heads twice?

This is an independent event because the first event, the coin toss, does not effect the second event, the second toss.

The probability of these two independent events is $$ \frac 1 4 $$ !

The sample space for these tosses illustrate the 4 distinct ways that the first toss followed by the second toss can play out.

First Toss

Second Toss

What if we flip the coin three times?

3 independent events

What is the probability that we will get heads exactly three times if we toss the coin three times?

The probability of getting heads all three times is $$ \frac 1 8 $$ .

Let's look at the sample space for these tosses:

Three ways that we can get 1 Heads out of 3 tosses

Three ways that we can get 2 heads out of 3 tosses

1 way to get 3 heads over 3 tosses

Developing the Formula

If we spin the red arrow 4 times. What is the probability that we will spin an odd number on the 1^{st} spin and even numbers on the final 3 spins?

p(odd)

p(even)

p(even)

p(even)

To calculate the probability of independent events simply multiply each probability together.

×
×
×
=
×
=

Extending the formula

Imagine that we are using the same spinner depicted up above. Calculate the probability of obtaining exactly 1 odd number on 4 spins of the arrow. This question is different because you can get an odd number any time. Maybe on the second spin .

1st spin

2nd spin

3rd spin

4th spin

Probability

p(odd)
$\frac 2 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(odd)
$\frac 2 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(odd)
$\frac 2 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(even)
$\frac 1 3 $

p(odd)
$\frac 2 3 $

So, remember we changed the question. We can have an odd number for any spin, maybe there's an odd number in the first spin or in the second spin. So we add each of the $$ \frac{2}{81}$$ probabilities up to get our answer:
Note, this is the same as
.

So if there are 4 possible outcomes and you want exactly 1 of them to occur, the formula is
$
_4 C _1 \cdot p(\text{ success})^1 \cdot p(\text{fail})^{(4-1)}
\\
_4 C _1 \cdot p(\text{ success})^1 \cdot p(\text{fail})^3
$

Formula = The general formula is to determine how many combinations of the independent events can occur, then multiply the probability of each by the result of the combination. It's much easier to understand by looking at some more examples.

Problem 1

What is the probability of obtaining two odd numbers on four spins?