Ad Clicks - Binomial Distribution
An advertisement is shown to 20 unique users. Each user clicks on the ad with an independent probability of 10%.
- What is the probability that exactly 2 users click on the ad?
- What is the probability that at least 1 user clicks on the ad?
Related Concepts
Hint
First, let's identify the key pieces of information for our Binomial distribution formula:
- Number of trials (n): How many users see the ad? (Answer: 20)
- Probability of success (p): What's the chance one user clicks? (Answer: 10% or 0.1)
- Probability of failure (1-p): What's the chance one user *doesn't* click? (Answer: 1 - 0.1 = 0.9)
- For "exactly 2 users click":
- Here, the number of successes (k) is 2.
- Use the Binomial formula:
P(X = k) = C(n, k) * pk * (1-p)n-k - You'll need to calculate C(20, 2). This means "how many ways can you choose 2 users out of 20?" The formula is
C(n, k) = n! / (k! * (n-k)!). So, C(20, 2) = 20! / (2! * 18!) = (20 × 19) / (2 × 1) = 190. - Then calculate pk which is (0.1)2.
- Then calculate (1-p)n-k which is (0.9)(20-2) = (0.9)18.
- Multiply these three parts together.
- For "at least 1 user clicks":
- "At least 1" means 1 click, OR 2 clicks, OR 3 clicks, ... all the way up to 20 clicks. Calculating all of these and adding them up would take a long time!
- There's a much easier way: think about the opposite. What's the only situation that's NOT "at least 1 click"? It's "zero clicks" (P(X = 0)).
- The probability of "at least 1 click" is
1 - (probability of 0 clicks). This is called the complement rule. - So, first calculate P(X = 0) using the Binomial formula (with k=0).
- C(20, 0) is 1 (there's only 1 way to choose 0 users out of 20).
- (0.1)0 is 1 (any number to the power of 0 is 1).
- Then subtract that result from 1.
Explanation: Ad Clicks - Binomial Distribution
Let's imagine we show an online ad to 20 different people. For each person, there's a 10% chance they will click on the ad. Each person decides to click or not click on their own, without being influenced by others.
This situation is perfect for the Binomial distribution because:
- We have a fixed number of tries (20 people, so 20 "trials").
- Each try has two outcomes: a person either "clicks" (success) or "does not click" (failure).
- The chance of success (clicking) is the same for everyone (10% or 0.1).
- Each person's decision is independent.
Now, let's answer the questions:
- What's the chance that exactly 2 people out of the 20 click the ad? To figure this out, we need to consider:
- The chance of 2 specific people clicking (0.1 × 0.1).
- The chance of the other 18 people NOT clicking (0.9 multiplied by itself 18 times).
- How many different groups of 2 people can we choose from the 20? (This is where "combinations" come in).
- What's the chance that at least 1 person clicks the ad? "At least 1" means it could be 1 person, or 2 people, or 3, ..., all the way up to 20 people clicking. Calculating each of those probabilities and adding them up would be a lot of work!
Instead, it's much easier to think about the opposite scenario: What's the chance that NO ONE clicks (i.e., 0 people click)?
If we find the chance of "0 clicks," then the chance of "at least 1 click" is simply 1 minus (the chance of 0 clicks). This is a handy trick called the complement rule.
Let X be the random variable representing the number of users who click the ad. We have:
- Number of trials (users shown the ad): n = 20.
- Probability of success (a user clicks): p = 0.1.
- Probability of failure (a user does not click): 1-p = 1 - 0.1 = 0.9.
Since these conditions fit the Binomial model, X ~ Binomial(n=20, p=0.1).
The formula for the Binomial probability is: P(X = k) = C(n, k) * pk * (1-p)n-k
1. Probability of exactly 2 users clicking (P(X = 2))
We want to find the probability when k = 2 successes.
Step 1: Calculate C(n, k) - the number of combinations. This tells us how many ways we can choose 2 users out of 20. C(n, k) = n! / (k! * (n-k)!) C(20, 2) = 20! / (2! * (20-2)!) C(20, 2) = 20! / (2! * 18!) To simplify this without calculating huge factorials: C(20, 2) = (20 × 19) / (2 × 1) C(20, 2) = 380 / 2 = 190. So, there are 190 different pairs of users who could be the ones clicking.
Step 2: Calculate pk. This is the probability of 'k' successes. pk = (0.1)2 = 0.1 × 0.1 = 0.01.
Step 3: Calculate (1-p)n-k. This is the probability of 'n-k' failures. Number of failures = n - k = 20 - 2 = 18. (1-p)n-k = (0.9)18. Using a calculator for (0.9)18: (0.9)18 ≈ 0.1500946 (We'll keep a few decimal places for now).
Step 4: Multiply all parts together. P(X = 2) = C(20, 2) × p2 × (1-p)18 P(X = 2) ≈ 190 × 0.01 × 0.1500946 P(X = 2) ≈ 1.9 × 0.1500946 P(X = 2) ≈ 0.28517974
Rounding to three decimal places, the probability of exactly 2 users clicking is approximately 0.285 (or about 28.5%).
2. Probability of at least 1 user clicking (P(X ≥ 1))
As discussed in the beginner explanation, it's easier to find the probability of the opposite event (0 users clicking) and subtract it from 1. P(X ≥ 1) = 1 - P(X = 0)
Step 1: Calculate P(X = 0) (probability of exactly 0 users clicking). Here, k = 0.
C(20, 0) = 20! / (0! * 20!) = 1. (There's only one way for no one to click).p0 = (0.1)0 = 1. (Any non-zero number to the power of 0 is 1).(1-p)n-0 = (0.9)20. Using a calculator for (0.9)20:(0.9)20 ≈ 0.12157665.
P(X = 0) = C(20, 0) × (0.1)0 × (0.9)20 P(X = 0) ≈ 1 × 1 × 0.12157665 P(X = 0) ≈ 0.12157665.
Step 2: Use the complement rule. P(X ≥ 1) = 1 - P(X = 0) P(X ≥ 1) ≈ 1 - 0.12157665 P(X ≥ 1) ≈ 0.87842335
Rounding to three decimal places, the probability of at least 1 user clicking is approximately 0.878 (or about 87.8%).
Summary of Results
- The probability that exactly 2 users click on the ad is approximately 0.285.
- The probability that at least 1 user clicks on the ad is approximately 0.878.
Note on Approximations: Sometimes, problems provide slightly rounded values for powers to simplify manual calculation. For example, if a problem stated (0.9)18 ≈ 0.1501, then using this value: P(X=2) ≈ 190 × 0.01 × 0.1501 = 1.9 × 0.1501 = 0.28519, which still rounds to 0.285. Similarly, if (0.9)20 ≈ 0.122 was given: P(X=0) ≈ 0.122, and P(X ≥ 1) = 1 - 0.122 = 0.878. Using a calculator for the powers directly provides more precision for intermediate steps. The final answers here are based on using more precise values for the powers and then rounding.
What's the Expected Number? On average, how many users would you expect to click the ad out of these 20 users? For a Binomial distribution, the expected number of successes (average number) is calculated very simply: E[X] = n × p. Can you calculate it?