Understanding Probability and Odds

The question asks for the probability that at least two out of three candidates solve a question correctly. We are given the odds in favour of each candidate solving the question. First, let's understand how to convert odds in favour into probability.

If the odds in favour of an event are \(a:b\), it means that for every \(a\) times the event happens, it does not happen \(b\) times. The total number of outcomes is \(a + b\). Therefore, the probability of the event happening is \(\frac{a}{a+b}\).

Converting Odds to Probabilities

Let \(C_1\), \(C_2\), and \(C_3\) represent the events that Candidate 1, Candidate 2, and Candidate 3 solve the question correctly, respectively.

• For Candidate 1, odds in favour are 5:2.
  Probability \(P(C_1) = \frac{5}{5+2} = \frac{5}{7}\).
  Probability of not solving \(P(C_1') = 1 - P(C_1) = 1 - \frac{5}{7} = \frac{2}{7}\).
• For Candidate 2, odds in favour are 4:3.
  Probability \(P(C_2) = \frac{4}{4+3} = \frac{4}{7}\).
  Probability of not solving \(P(C_2') = 1 - P(C_2) = 1 - \frac{4}{7} = \frac{3}{7}\).
• For Candidate 3, odds in favour are 3:4.
  Probability \(P(C_3) = \frac{3}{3+4} = \frac{3}{7}\).
  Probability of not solving \(P(C_3') = 1 - P(C_3) = 1 - \frac{3}{7} = \frac{4}{7}\).

The probabilities are summarized in the table below:

Calculating Probability of At Least Two Correct Solutions

The event "at least two of them solve the question correctly" means either:

• Exactly two candidates solve the question correctly, OR
• Exactly three candidates solve the question correctly.

Since these two cases are mutually exclusive (they cannot happen at the same time), the probability of "at least two solving correctly" is the sum of the probabilities of these two cases.

Case 1: Exactly Two Candidates Solve Correctly

There are three possibilities for exactly two candidates solving correctly:

• Candidate 1 solves, Candidate 2 solves, and Candidate 3 does not solve (\(C_1 \text{ and } C_2 \text{ and } C_3'\)).
  Probability \(P(C_1 \cap C_2 \cap C_3') = P(C_1) \times P(C_2) \times P(C_3')\) (since the events are independent).
  \(P(C_1 \cap C_2 \cap C_3') = \frac{5}{7} \times \frac{4}{7} \times \frac{4}{7} = \frac{5 \times 4 \times 4}{7 \times 7 \times 7} = \frac{80}{343}\).
• Candidate 1 solves, Candidate 2 does not solve, and Candidate 3 solves (\(C_1 \text{ and } C_2' \text{ and } C_3\)).
  Probability \(P(C_1 \cap C_2' \cap C_3) = P(C_1) \times P(C_2') \times P(C_3) = \frac{5}{7} \times \frac{3}{7} \times \frac{3}{7} = \frac{5 \times 3 \times 3}{7 \times 7 \times 7} = \frac{45}{343}\).
• Candidate 1 does not solve, Candidate 2 solves, and Candidate 3 solves (\(C_1' \text{ and } C_2 \text{ and } C_3\)).
  Probability \(P(C_1' \cap C_2 \cap C_3) = P(C_1') \times P(C_2) \times P(C_3) = \frac{2}{7} \times \frac{4}{7} \times \frac{3}{7} = \frac{2 \times 4 \times 3}{7 \times 7 \times 7} = \frac{24}{343}\).

The total probability of exactly two candidates solving correctly is the sum of these probabilities:
Probability (Exactly Two Solve) = \(\frac{80}{343} + \frac{45}{343} + \frac{24}{343} = \frac{80 + 45 + 24}{343} = \frac{149}{343}\).

Case 2: Exactly Three Candidates Solve Correctly

This happens when Candidate 1 solves, Candidate 2 solves, and Candidate 3 solves (\(C_1 \text{ and } C_2 \text{ and } C_3\)).
Probability \(P(C_1 \cap C_2 \cap C_3) = P(C_1) \times P(C_2) \times P(C_3)\) (since the events are independent).
\(P(C_1 \cap C_2 \cap C_3) = \frac{5}{7} \times \frac{4}{7} \times \frac{3}{7} = \frac{5 \times 4 \times 3}{7 \times 7 \times 7} = \frac{60}{343}\).

Total Probability for At Least Two Solving Correctly

The probability that at least two of them solve the question correctly is the sum of the probabilities of Case 1 and Case 2:
Probability (At Least Two Solve) = Probability (Exactly Two Solve) + Probability (Exactly Three Solve)
Probability (At Least Two Solve) = \(\frac{149}{343} + \frac{60}{343} = \frac{149 + 60}{343} = \frac{209}{343}\).

Therefore, the probability that at least two of the candidates solve the question correctly is \(\frac{209}{343}\).

Revision Table: Probability Concepts

Additional Information: Odds vs. Probability

It's important to distinguish between odds and probability. Probability is a ratio comparing the number of favourable outcomes to the total number of possible outcomes (favourable + unfavourable). Odds, on the other hand, compare the number of favourable outcomes to the number of unfavourable outcomes.

• Probability: success / (success + failure)
• Odds in Favour: success : failure
• Odds Against: failure : success

Knowing how to convert between odds and probability is crucial for solving problems like this one. If you are given odds against \(b:a\), the odds in favour are \(a:b\), and the probability is still \(\frac{a}{a+b}\).

In this problem, the candidates solving the question are treated as independent events. This means that whether one candidate solves the question or not does not influence whether another candidate solves it. This independence allows us to multiply their individual probabilities to find the probability of combined outcomes (like \(C_1\) solves AND \(C_2\) solves AND \(C_3\) doesn't solve).