Example 1

Solution 1 - Use web app

The app link is available from https://gallery.shinyapps.io/dist_calc/

Solution 2 - Use R dbinom()

Alternatively, using R dbinom() function, we set the number of successes, the first argument to 6, the size to 10 and the probability of success to 0.56, and obtain the same answer 24.3%.

> dbinom(6, size = 10, p = 0.56)
 [1] 0.243

Solution 3 - Calculate by hand

We are looking for the probability of exactly 6 successes in 10 trials.

P(k=6) =  (10 6) × 0.566 × 0.444
 
            10!
      = ---------- × 0.566 × 0.444
          6! × 4!
 
     = 210 × 0.566 × 0.444
 
     = 0.243

Example 2

Solution 1 - use the law of large numbers

The probability of success is P = 0.56.

In example 1:

• The sample size is 10;
• The expected number of successes would be 10 × 0.56 = 5.6.
• The difference between the expected and the desired number is then would be Δ = 6 - 5.6 = 0.4;

In this example:

• The sameple size is 1000;
• The expected number of successes is 1000 × 0.56 = 560;
• The difference between the expected and the desired number is Δ = 600 - 560 = 40;

Here, the desired outcome is much farther than the expected outcome and based on the law of large numbers, obtaining 600 successes when the expected is 560 should be a much less likely outcome that obtaining 6 successes when the expected is 5.6. Therefore, the answer is something less than 0.243.

Solution 2 - use R

Alternatively, using R dbinom() function, we set the number of successes, the first argument to 600, the size to 1000 and the probability of success to 0.56, and obtain the same answer 24.3%.

> dbinom(600, size = 1000, p = 0.56)
 [1] 0.00098

Example 3

The probability of success is P = 0.56, as mentioned in Example 1. This time n = 100, this seems like a larger sample size. So let's see if it is actually large enough to yield a nearly normal distribution.

The rule is that we need at least 10 expected successes and 10 expected failures.

Success: np = 100 × 0.56 = 56 > 10
Fail: n(1-p) = 100 × 0.44 = 44 > 10

Therefore, we know the shape of this distribution will be nearly normal. Normal distribution has two parameters, mean and standard deviation. So to fully describe the distribution, we need to calculate these parameters, which we know can be estimated by the binomial, mean and standard devition.

μ = 56
σ = sqrt(100 × 0.56 × 0.44) = 4.96

Example 4:

Solution 1 - Use web app

The app link is available from https://gallery.shinyapps.io/dist_calc/

Solution 2 - Use R dbinom() and sum()

Alternatively, using R dbinom() function to get the probability and sum() function to get the bunch of probabilities. We set the number of successes, the first argument to 60:100, the size to 100 and the probability of success to 0.56, and obtain the same answer 24.3%.

> sum(dbinom(60:100, size = 100, p = 0.56))
 [1] 0.241

Solution 3 - Work out using normal distribution

We can make use of the fact that the distribution is nearly normal, with mean 56 and standard deviation 4.96.

When finding normal probabilities, we calculate the z score as the observation.

      60 - 56
Z = ----------- = 0.81
        4.96

Next step is to find the probability of Z > 0.81 from the normal probability table, that is roughly 0.209.

P(Z > 0.81) = 0.209

Once again, this probability is a little lower than the probabilities we have calculated using the App and R. This discrepancy is mostly due to the fact that under the normal distribution, probability of exactly 60 successes is undefined. To account for that, we apply a 0.5 correction to the observation of interest, and can work through the problem.

      59.5 - 56
Z = ------------- = 0.71 
        4.96

Then the updated probability is:

P(Z > 0.71) = 0.239 

The result is much closer to the exact probability.

References & Resources;

• N/A