Probability of Obtaining at Least 2 Heads in Five Coin Tosses
When tossing a fair coin multiple times, determining the probability of obtaining a specific result can often be complex. In this article, we will explore the probability of obtaining at least 2 heads in five coin tosses. We will cover different methods to calculate this probability and provide a detailed explanation for each approach.
Introduction to Probability in Coin Tossing
Each coin toss has two possible outcomes: heads (H) or tails (T). When a fair coin is tossed, the probability of each outcome is 0.5. With five coin tosses, there are a total of (2^5 32) possible outcomes. Let's explore how to calculate the probability of obtaining at least 2 heads in these 32 outcomes.
Method 1: Counting the Favorable Outcomes
One way to determine the probability is by counting the number of favorable outcomes and dividing it by the total number of possible outcomes. Let's break this down:
Total number of outcomes: [2^5 32]
Favorable outcomes: We need to count the number of outcomes that have at least 2 heads. These outcomes include having exactly 2, 3, 4, or 5 heads.
Counting outcomes with at least 2 heads:
Exactly 2 heads: (binom{5}{2} 10) Exactly 3 heads: (binom{5}{3} 10) Exactly 4 heads: (binom{5}{4} 5) Exactly 5 heads: (binom{5}{5} 1)Total number of favorable outcomes: (10 10 5 1 26)
Probability: [frac{26}{32} frac{13}{16} 0.8125]
Method 2: Complementary Counting
Another approach is to use complementary counting. This involves calculating the probability of the complementary event (i.e., the event that we do not want to happen) and subtracting it from 1.
Probability of no heads: [left(frac{1}{2}right)^5 frac{1}{32}]
Probability of exactly one head: There are 5 positions for the head, and each position has a probability of (frac{1}{2^5} frac{1}{32}). Therefore, the total probability is: [binom{5}{1} left(frac{1}{2}right)^5 5 times frac{1}{32} frac{5}{32}]
Total probability of no heads or exactly one head: [frac{1}{32} frac{5}{32} frac{6}{32} frac{3}{16}]
Probability of at least 2 heads: [1 - frac{3}{16} frac{13}{16} 0.8125]
Method 3: Generalized Approach for Any Number of Tosses
Let's generalize this to find the probability of obtaining at least 2 heads when tossing (n) coins. The total number of outcomes is (2^n). The number of outcomes with no heads or exactly one head is (binom{n}{0} binom{n}{1} 1 n). Therefore, the probability of obtaining at least 2 heads is:
[P(text{at least 2 heads}) 1 - left(frac{1 n}{2^n}right)]For (n 5), we get:
[P(text{at least 2 heads}) 1 - left(frac{1 5}{2^5}right) 1 - left(frac{6}{32}right) frac{26}{32} frac{13}{16} 0.8125]Conclusion
In conclusion, the probability of obtaining at least 2 heads in five coin tosses is (frac{13}{16}) or 0.8125. This value can be derived through various methods, including counting favorable outcomes, complementary counting, and a generalized approach. Understanding these methods can help in solving similar probability problems involving binomial distributions and multiple trials.