Basic probability exercise

Question

Consider the problem of selecting two candidates from a group of five persons for a job. Knowing that the candidates differ in their degree of readiness (1 is the best prepared, 2 is less prepared than 1, and so on to the fifth, the worst prepared). The chief of staff, of course, knows nothing of this classification. Find the probabilities of the following events:

1. The chief of staff selects the best and one of the two worst candidates.
2. The chief of staff selects at least one of the two best.

Then, the sample space is: $S=\lbrace 1,2,3,4,5 \rbrace$

So, my questions:

Should I assume that the first and the second candidates are chosen simultaneously?, if so:

1. $S=\lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),...,(4,5),(5,5) \rbrace$

$P(A)={2\over{25}}$ taking $A=\lbrace (1,5), (5,1)\rbrace \subset S$

and

$P(B)={2\over25}$ with $B= \lbrace (2,5), (5,2) \rbrace \subset S$

Then the probability is: $P(A) + P(B) = 0,16$

Or should I consider when selecting the first, the sample space is reduced?, this is:

$P_1=P(X=5)={1\over{5}}$, and now the space is only $S=\lbrace 1,2,3,4 \rbrace$, so $P_2=P(X=1 \lor X=2)=P(X=1)+P(X=2)={1\over{4}}+{1\over{4}}=0.5$

So the probability is $P_1 \cdot P_2 = 0,2 \cdot 0,5=.1$

What is the right solution?

Thanks!

Answer 1

As mentioned in the comments above, presumably the chief of staff will pick two different people (implied by the phrasing $\underline{\text{selecting two candidates}}$).

This becomes then a question of selecting two people without repetition. We may assume that order is unimportant or that order is important. Both will arrive at the same answer. I personally prefer to work with order being unimportant, so that is how I will proceed here.

Our sample space will be all ways of selecting two people from the five where order doesn't matter. This corresponds to the set of subsets of size two of $[5]$, namely the set $\{\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\}\}$ (remember $\{2,3\}$ is the same as $\{3,2\}$ in this construction) The sample space in this construction will be of size $\binom{5}{2}=10$

Each of these will be equiprobable, and hence we may use the definition of probability in an equiprobable space being $\frac{\text{number of good combinations}}{\text{number of combinations regardless}}$

To proceed, since this is a relatively small example, we could simply look at the list of possibilities and count by hand how many satisfy each condition. This is not helpful for learning how to do the same problem with a larger group however, and you run the risk of counting incorrectly. Instead, let us use a combinatorial argument to count.

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In how many ways can we pick two people such that one of them is labeled $1$ and the other is labeled either $4$ or $5$? (where order doesn't matter)

The only missing piece of information is whether the low ranking person is either the four or the five. There are then two possibilities.

What is the probability then that the chief of staff picks the best and one of the two worst people?

$\frac{2}{10}=0.2$

In how many ways can the chief of staff pick at least one of the two best?

He could have picked both the $1$ and the $2$, the $1$ and something else other than a $2$, or a $2$ and something else other than a $1$. This gives a total of $1+3+3=7$ possibilities.

Alternatively, we may approach using inclusion-exclusion. The number of ways to pick a $1$ and something else plus the number of ways to pick a $2$ and something else minus the number of ways to pick a $1$ and $2$ simultaneously. $4+4-1=7$

The probability of this occuring is then:

$\frac{7}{10}=0.7$

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In the case that you consider order important, the sample space will then be $\{(1,2),(1,3),\dots,(2,1),(2,3),\dots\}$ and will be of size $5\cdot 4=20$

The number of possibilities for the first question will be $2\cdot 2=4$ seen by multiplication principle:

• select whether the person labeled $1$ is first picked or second picked
• select which of the two worst people picked

The probability is then $\frac{4}{20}=0.2$, same as before.

Similarly, the probability of the second question will be $\frac{14}{20}=0.7$, seen by multiplication/addition principle:

• case 1: first person picked is one of the two best. Pick which one it is. $2$ options
  • pick who the second person picked is then. $4$ options
• case 2: first person picked is not one of the two best. Pick which one it is. $3$ options
  • pick who the second person picked is then. It must be one of the two best. $2$ options

For a total of $2\cdot 4+ 3\cdot 2 = 8+6=14$ possibilities

Answer 2

There are $\binom{5}{2}$ possibilities: selection without repeatition, order irrelevant. $$S=\{(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)\}$$

There are two ways to select the best and one of the two worst: $A=\{(1,4), (1,5)\}$

There are $7$ ways to select at least one of the two best: either select just one of them and one of the three others, or select both of them. $$\lvert B\rvert=\binom{2}{1}\binom{3}{1}+\binom{2}{2}$$