Question

Consider a rat going through a maze, and each time the rat begins the maze he has 30% chance of finishing successfully. The rat goes through the maze over and over again until he is successful in finishing the maze. Whether or not the rat finishes the maze on one trial has no impact on his chance of finishing the maze on the next trial.
What is the probability that the rat fails the maze 6 times in a row, and then succeeds on his 7th attempt? Round your answer to two decimal places.

Answer 1

Answer:

3.53% probability that the rat fails the maze 6 times in a row, and then succeeds on his 7th attempt

Step-by-step explanation:

For each time that the rat goes through the maze, there are only two possible outcomes. Either he fails it, or he succeds. The trials are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

What is the probability that the rat fails the maze 6 times in a row, and then succeeds on his 7th attempt?

Missing on the first six, each with 70% probability(P(X = 6) when p = 0.7, n = 6).

Succeeding on the 7th attempt, with p = 0.3. So

$P = 0.3P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.7)^{6}.(0.3)^{0} = 0.117649$

$P = 0.3P(X = 6) = 0.3*0.117649 = 0.0353$

3.53% probability that the rat fails the maze 6 times in a row, and then succeeds on his 7th attempt