Understanding the Probability of Tossing Four Coins and Getting Exactly Two Heads

The concept of probability is fundamental in understanding the outcome of random events, such as tossing coins. This article will explore how to calculate the probability of obtaining exactly two heads when tossing four coins, using the binomial probability formula. We will also provide a practical demonstration to enhance your understanding of the concept.

Binomial Probability Formula

In the context of probability, the binomial distribution is used to model the number of successes in a fixed number of independent Bernoulli trials. For this example, we are dealing with four independent coin tosses, each with two possible outcomes: heads (H) or tails (T). The probability of success (getting heads) on a single toss is ( p frac{1}{2} ).

Formula Explanation

The binomial probability formula is given by:

P(X) C(n, k) * p^k * (1-p)^{n-k}

n - the total number of trials (coin tosses) in this case, ( n 4 ). k - the number of successful outcomes (heads in this case) we are interested in, which is ( k 2 ). p - the probability of getting heads on a single toss, ( p frac{1}{2} ). C(n, k) - the binomial coefficient, which calculates the number of ways to choose ( k ) successes in ( n ) trials.

Calculating Binomial Coefficient

First, we calculate the binomial coefficient ( C(4, 2) ) using the formula:

C(n, k) n! / (k! * (n-k)!)

Plugging in the values:

C(4, 2) frac{4!}{2! * (4-2)!} frac{4 * 3}{2 * 1} 6

Applying the Binomial Probability Formula

Now, we substitute the values into the binomial probability formula:

P(X 2) C(4, 2) * left(frac{1}{2}right)^2 * left(frac{1}{2}right)^{4-2}

P(X 2) 6 * left(frac{1}{2}right)^2 * left(frac{1}{2}right)^2

P(X 2) 6 * left(frac{1}{2}right)^4 6 * frac{1}{16} frac{6}{16} frac{3}{8}

Therefore, the probability of tossing four coins and getting exactly two heads is (frac{3}{8}), or (0.375).

Visualizing the Outcomes

To further illustrate, let's list all possible outcomes when tossing four coins:

2211 1221 1122 2121 1212 2112 1112 1121 2221 2122 1222 1111 1211 2111 1121 2212

Out of the 16 possible outcomes, 6 of them result in exactly 2 heads. This confirms our calculated probability of ( frac{3}{8} ).

Additional Probabilities

Let's explore other possible scenarios:

Three Heads and One Tail

The probability of obtaining three heads and one tail is ( frac{4}{16} ), as there are 4 outcomes that satisfy this condition:

1222 2122 2212 2221

Four Heads, No Tails

The probability of getting four heads is ( frac{1}{16} ), as there is only one outcome that satisfies this condition:

2222

Adding these probabilities together, we get:

(frac{6}{16} frac{4}{16} frac{1}{16} frac{11}{16} 0.6875)

This means there is a 68.75% chance of getting at least two heads when tossing four coins.

Understanding the binomial probability formula and its application not only helps in solving theoretical problems but also provides insight into real-world scenarios like gambling, statistics, and data analysis.