# Conditional probability Maths question

The following questions are an excerpt from a set of questions that are to be completed as preparation to my exam on Friday. Myself and friends believe the answer to part 1i to be 1/7 or 0.143 and the answer for part 1ii to be 0.185. However, the problem myself and friends have is that the mark scheme provided by our teacher states a different answer to part 1i. For the second question, too, we are also having difficulty ascertaining the correct answer. If anyone could spare a couple of minutes to work out the answers to both so that we can compare answers, that would be appreciated massively.

1. A man has 7 keys, of which only one will open the door of his home.

(i) Suppose he tries the keys at random, discarding (i.e. setting aside) those that do not work. What is the probability that he will open the door on the 5th try?

(ii) Suppose he tries the keys at random without discarding those that do not work. What is the probability he will open the door on the 5th try?

1. Four Americans, 3 Frenchmen, and 3 Englishmen are to be seated in a row. How many seating arrangements are possible when people of the same nationality must sit next to each other?

For the first key question, the simplest approach is as follows. All sequences of keys are equally likely. So the probability the right key is fifth is $\frac{1}{7}$.

For the second question, a randomly chosen key has probability $\frac{1}{7}$ of success, and probability $\frac{6}{7}$ of failure. The experiment is repeated independently several times. The probability of $4$ failures in a row followed by success is $\left(\frac{6}{7}\right)^4 \cdot\frac{1}{7}$.

For the seating, the nations can be ordered in $3!$ ways. For each such way, the individuals can be ordered within national groupings in $4!3!3!$ ways, for a total of $3!4!3!3!$.

Remark: There are longer ways to find the first answer. For example, the probability the first key is wrong is $\frac{6}{7}$. Given the first key is wrong, the probability the second key is wrong is $\frac{5}{6}$. And so on. finally, given that the first $4$ are wrong, the probability the fifth is right is $\frac{1}{3}$. The required probability is therefore $\frac{6}{7}\cdot\frac{5}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}$. This simplifies to $\frac{1}{7}$.