Calculating Probabilities: Steak and Salad Orders

This article delves into the methodology of calculating the probability of a specific event using combinatorics and probability concepts. The context here is ten people ordering either steak or salad, each with a probability of 1/2. The main objective is to determine the probability that one more person orders steak than salad.

Introduction

The problem at hand involves determining the probability that the number of people ordering steak is one greater than the number who order salad. With ten individuals involved, each has a 50% chance of ordering steak or salad. This section outlines the steps to solve such a probability problem.

Understanding the Problem

The scenario involves ten people each having a 1/2 probability of ordering steak or salad. We need to find the probability that one more person orders steak than salad. This can be approached using the binomial distribution, where the number of trials (people) is 10 and the probability of success (ordering steak) is 1/2.

Brute-Force Method

A straightforward way to calculate this probability is by listing out all the possible scenarios where the number of steak orders is exactly one more than the number of salad orders. Each person can either order steak or salad, leading to a total of (2^{10}) possible orders. We can then count the number of favorable outcomes and divide by the total number of outcomes.

For example, the possibilities are as follows:

10st9sa9st8sa...1st0sa

There are 10 different ways that this can happen, and there are (2^{10} 1024) total possible outcomes. Therefore, the probability is:

(P frac{10}{1024} ≈ 0.009765625)

Using Combinatorics

To find the probability using combinatorics, we define the probability (P_r) that exactly (r) people order an item. The probability of exactly (r) out of 10 people ordering a specific item (steak or salad) can be calculated using the binomial coefficient ({_{10}text{C}_r}).

The formula for the number of combinations is:

(_{10}text{C}_r frac{10!}{r!(10 - r)!})

We can also use the property that (_{10}text{C}_r _{10}text{C}_{10-r}).

Step-by-Step Calculation

Let's calculate the combinations for each possible value of (r):

_{10}text{C}_0  1
_{10}text{C}_1  10
_{10}text{C}_2  45
_{10}text{C}_3  120
_{10}text{C}_4  210
_{10}text{C}_5  252
_{10}text{C}_6  210
_{10}text{C}_7  120
_{10}text{C}_8  45
_{10}text{C}_9  10
_{10}text{C}_{10}  1

The total number of possible outcomes is (2^{10} 1024).

Now, let's calculate the probability (P) that the number of steak orders is exactly one more than the number of salad orders:

(P frac{1}{2^{10}} left[,1_{_{10}text{C}_0} 10_{_{10}text{C}_1} 45_{_{10}text{C}_2} 120_{_{10}text{C}_3} 210_{_{10}text{C}_4} 252_{_{10}text{C}_5} 210_{_{10}text{C}_6} 120_{_{10}text{C}_7} 45_{_{10}text{C}_8} 10_{_{10}text{C}_9} 1_{_{10}text{C}_{10}} right])

This simplifies to:

(P frac{10 450 5400 25200 52920}{524288} ≈ 0.16018)

Therefore, the probability that the number of steak orders is exactly one more than the number of salad orders is approximately 0.16018.

Conclusion

This problem involves understanding combinatorics and probability concepts, particularly the binomial distribution. By breaking down the problem into manageable steps and using the binomial coefficient, we can calculate the probability of specific outcomes. The detailed calculation above demonstrates the process and provides a clear solution to the problem.