In a race, the odds in favour of cars P, Q, R and S winning are given respectively as 1:3, 1:4, 1:5 and 1:6. Assuming these odds correctly represent their chances against all other competitors, what is the probability that one of these four cars (P, Q, R or S) wins the race?

Introduction / Context:
This problem involves interpreting odds in favour of different cars winning a race and converting those odds into probabilities. Four cars P, Q, R and S each have odds in favour of winning given against all other competitors. We must find the probability that one of these four cars wins the race, which requires converting each set of odds into a probability and then adding them together.

Given Data / Assumptions:

• Odds in favour of car P winning are 1:3.
• Odds in favour of car Q winning are 1:4.
• Odds in favour of car R winning are 1:5.
• Odds in favour of car S winning are 1:6.
• Odds in favour a:b mean probability of event = a / (a + b).
• We assume the events of each car winning are mutually exclusive and the odds are consistent with an underlying probability model.

Concept / Approach:
The main idea is to convert odds in favour of each car into probabilities. For odds a:b in favour, the probability that the event occurs is a / (a + b). We do this for each car, obtaining P(P wins), P(Q wins), P(R wins) and P(S wins). Because only one car can win the race, and the events are mutually exclusive, the probability that any one of the four wins is the sum of these four probabilities.

Step-by-Step Solution:
Step 1: For odds 1:3 in favour of P, the probability P(P wins) = 1 / (1 + 3) = 1 / 4.Step 2: For odds 1:4 in favour of Q, the probability P(Q wins) = 1 / (1 + 4) = 1 / 5.Step 3: For odds 1:5 in favour of R, the probability P(R wins) = 1 / (1 + 5) = 1 / 6.Step 4: For odds 1:6 in favour of S, the probability P(S wins) = 1 / (1 + 6) = 1 / 7.Step 5: The probability that one of these four cars wins is the sum: 1/4 + 1/5 + 1/6 + 1/7.Step 6: Compute this sum using a common denominator. The least common multiple of 4, 5, 6 and 7 is 420. The sum becomes (105 + 84 + 70 + 60) / 420 = 319 / 420.
Verification / Alternative check:
We can verify the arithmetic by checking each term carefully. Converting to denominator 420 gives 1/4 = 105/420, 1/5 = 84/420, 1/6 = 70/420 and 1/7 = 60/420. Adding these numerators gives 105 + 84 + 70 + 60 = 319, so the total probability is 319/420. This is less than 1, as it must be, which is sensible since the remaining probability accounts for all other competitors in the race who are not among P, Q, R and S.

Why Other Options Are Wrong:
27/111 is an oddly simplified fraction that does not result from the natural denominators involved in the odds. 114/121 is greater than 1, which is impossible for a probability. 231/420 would correspond to the sum of only some of the component probabilities and is therefore too small. 1/4 is simply the probability for car P alone and does not include the contributions from Q, R and S.

Common Pitfalls:
Students frequently misinterpret odds as probabilities, writing 1:3 as 1/3 when the actual probability is 1/4. It is also easy to forget to sum all four probabilities or to ignore that the events are mutually exclusive. Another common mistake is to simplify incorrectly when using a common denominator. Keeping track of each term and using a systematic method to add fractions avoids these errors.

Final Answer:
The probability that one of the cars P, Q, R or S wins the race is 319/420.