Data Interview Question

Coupon Utilization Probability

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Solution & Explanation

To solve the problem of determining the probability that at least one coupon will be redeemed by the riders, we need to delve into the principles of probability, specifically the binomial distribution. Here’s a step-by-step breakdown:

Understanding the Problem

  • N riders each receive a discount coupon.
  • Each coupon has a usage probability P.
  • We need to find the probability that at least one coupon is used.

Approach

The most effective way to solve this is to consider the complementary event: the probability that no coupons are used at all. Once we have this, we can subtract it from 1 to find our desired probability.

Complementary Probability

  • Probability that a single rider does not use their coupon:
    • This is simply (1 - P), given that P is the probability of using the coupon.
  • Probability that none of the N riders use their coupons:
    • For N riders, this probability is (1 - P)^N.

Calculating the Desired Probability

  • Probability that at least one coupon is used:

    • This is the complement of the probability that no coupons are used.
    • Therefore, it is given by:

    P(at least one coupon is used)=1(1P)NP(\text{at least one coupon is used}) = 1 - (1-P)^N

Explanation with an Example

Imagine there are 5 riders (N = 5), and each has a probability of 0.2 (P = 0.2) of using their coupon.

  1. Probability that a single rider does not use their coupon:
    • (1 - P) = 0.8
  2. Probability that none of the 5 riders use their coupons:
    • (1 - P)^5 = 0.8^5 ≈ 0.32768
  3. Probability that at least one coupon is used:
    • 1 - (1-P)^5 = 1 - 0.32768 ≈ 0.67232

Thus, there is approximately a 67.23% chance that at least one coupon will be redeemed.

Conclusion

The solution effectively uses the concept of complementary probability to simplify the computation. By focusing on the likelihood that no coupons are used and subtracting this from 1, the calculation becomes straightforward and intuitive. This approach is both efficient and insightful, making it a valuable method in scenarios involving multiple independent trials.