The problem can be solved using the Binomial Probability Formula:

where X is the number of successes, n is the number of times an event can occur, and p is the probability of success.
From the question, we can get the following parameters:

We can have that:

We have that:

Using the steps above, we can calculate the other values to be:

Therefore, the probability is calculated to be:

The probability is about 0.8023.