What is the probability that a random student would need exactly three tries to pass their driving test, given that 40% pass on the first try and 80% on subsequent attempts?

To determine the probability that a random student would need exactly three tries to pass their driving test, we can analyze the situation using the probabilities given.

The scenario dictates that the student needs to fail the first two attempts and then pass on the third attempt.

1. The probability of passing on the first attempt is 40%, so the probability of failing the first attempt is 60% (1 - 0.4 = 0.6).

2. For the second attempt, since they failed the first, they now have an 80% chance of passing, which means a 20% chance of failing (1 - 0.8 = 0.2).

3. Finally, on the third attempt, they again have an 80% chance of passing.

Now, to find the probability of needing exactly three tries, we multiply the probabilities of failing the first two attempts and passing on the third:

• Probability of failing the first try: 0.6

• Probability of failing the second try: 0.2

• Probability of passing on the third try: 0.8

Thus, the calculation is:

[

P(\text{3 tries}) = P(\text{Fail 1st})