We are dealing with a problem where six candidates $(A, B, C, D, E, F)$ can be selected to be on a show, and we are to calculate the probabilities of three specific events involving different numbers of selected candidates.

Total Number of Possible Subsets

For $n = 6$ candidates, each candidate can either be selected or not, leading to a total of: $2^n = 2^6 = 64$ possible subsets (including the empty subset where no candidate is selected).

Part (a): Event $E_1$: "Four candidates get selected"

To calculate the probability of selecting exactly 4 candidates, we need to determine how many ways we can select 4 candidates out of 6.

This is a combination problem, and the number of ways to choose 4 candidates from 6 is: $\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15$

Thus, the probability for $E_1$ is: $P(E_1) = \frac{\binom{6}{4}}{2^6} = \frac{15}{64}$

Part (b): Event $E_2$: "Five candidates get selected"

Similarly, the number of ways to choose 5 candidates from 6 is: $\binom{6}{5} = \frac{6!}{5!(6-5)!} = 6$

The probability for $E_2$ is: $P(E_2) = \frac{\binom{6}{5}}{2^6} = \frac{6}{64} = \frac{3}{32}$

Part (c): Event $E_3$: "Three candidates get selected, and $B$ is not one of them"

For this event, we are selecting 3 candidates out of the 5 remaining candidates (since $B$ is excluded).

The number of ways to select 3 candidates from the remaining 5 candidates is: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$

The probability for $E_3$ is: $P(E_3) = \frac{\binom{5}{3}}{2^6} = \frac{10}{64} = \frac{5}{32}$

Summary of Results:

• (a) $P(E_1) = \frac{15}{64}$
• (b) $P(E_2) = \frac{3}{32}$
• (c) $P(E_3) = \frac{5}{32}$

Would you like further details or explanations?

Here are five related questions you can explore next:

1. What is the probability that no candidate gets selected?
2. What is the probability that all six candidates get selected?
3. How would the probabilities change if there were 7 candidates instead of 6?
4. What is the probability of selecting exactly 2 candidates, given that $A$ is one of them?
5. What is the probability of selecting exactly 3 candidates, given that $A$ and $B$ must both be included?

Tip: In combinatorics, using combinations $\binom{n}{k}$ is a powerful tool to calculate selections from a set of $n$ items, where order doesn't matter.