# ok I get it, so once I get the prob figred for one person, how do I figure it if additional attempts are made and the question is about the prob of at least one of the additional attempts say n = 6 getting the same results as the first

Once you have the probability of passing the test with an >80% score happening once, call that P... It will be a very low number if t/f questions were guessed at random. The probability of not passing when trying once is 1-P (which is close to 1). . The probability of not passing after n independent attempts is (1-P)^n. The probability of passing after n attempts is
1 - (1-P)^n.

## To calculate the probability of at least one of the additional attempts (n = 6) getting the same results as the first, you can follow these steps:

Step 1: Calculate the probability of not passing the test after one attempt. As you mentioned, this is denoted as (1 - P), where P is the probability of passing with an >80% score.

Step 2: Calculate the probability of not passing after n independent attempts. This is done by raising the probability from step 1 to the power of n, denoted as (1 - P)^n. This gives you the probability of failing all n attempts.

Step 3: Finally, calculate the probability of passing after n attempts by subtracting the probability of not passing (from step 2) from 1. This is denoted as 1 - (1 - P)^n. This will give you the probability of passing at least once out of the n attempts.

For example, if the probability of passing with an >80% score (P) is 0.1, and you have n = 6 additional attempts, you would calculate it as follows:

Step 1: Probability of not passing after one attempt = 1 - P = 1 - 0.1 = 0.9.

Step 2: Probability of not passing after 6 independent attempts = (1 - P)^n = 0.9^6 = 0.531441.

Step 3: Probability of passing after 6 attempts = 1 - (1 - P)^n = 1 - 0.531441 = 0.468559.

Therefore, there is approximately a 46.86% chance of passing at least once out of the six additional attempts.