The possible outcomes are: AB, AC, AD, BC, BD, CD.

[There are a few explanations for each answer - hopefully at least one of them makes sense!]


Part (a)

Explanation 1: The probability is

MATH

since when we choose A, we must choose one of the remaining 3 directors to go with A. There are $C_{2}^{4}=6$ possible combinations.

Explanation 2: Probability that A is selected is MATH

[Choose A: $C_{1}^{1}$, and then choose one from the 3 remaining directors ($C_{1}^{3}$), divided by the number of possible outcomes: $C_{2}^{4}$.]


Part (b)

Explanation 1: The probability of getting A or B first is MATH.

Now to consider the probability of selecting A or B as the second director. In this case, the first director has to be C or D with probability $\dfrac{2}{4}$ (2 particular directors out of 4 possible).

Then the probability of the second being A or B is $\dfrac{2}{3}$ (2 particular directors out of the remaining 3 directors).

We need to multiply the two probabilities.

So the probability of getting A or B for the second director is MATH

The total is: MATH


Explanation 2: Probability that A or B is selected is

MATH

[Choose A as above, then choose B from the remaining 2 directors in a similar way.]


Explanation 3: If A or B is chosen, then we cannot have the case C and D is chosen. So the probability of A or B is given by:

$P($A or B$)=1-P($C and DMATH $\qquad $

$\qquad \qquad $

Part (c)

Probability that A is not selected is MATH


Extension: Consider the case if we are choosing 2 directors from 5. The probabilities would now be:

(a) Probability that A is selected is MATH

[Choose A: $C_{1}^{1}$, and then choose one from the 4 remaining directors ($C_{1}^{4}$), divided by the number of possible outcomes: $C_{2}^{5}$.]

(b) Probability that A or B is selected is

MATH

[Choose A as above and then choose B from the remaining 3].$\qquad \qquad $

(c) Probability that A is not selected is MATH.