Calculating Probability of At Least 6 Heads from 8 Coin Tosses

This problem involves calculating the probability of a specific outcome when tossing multiple coins simultaneously. Since each coin toss has only two possible outcomes (heads or tails) and the tosses are independent events, this scenario fits the characteristics of a binomial probability distribution.

We are tossing 8 coins simultaneously. We want to find the probability of getting at least 6 heads. This means we are interested in the events where we get exactly 6 heads, exactly 7 heads, or exactly 8 heads.

Let \(n\) be the number of coin tosses, so \(n = 8\). Let \(p\) be the probability of getting a head in a single toss. For a fair coin, \(p = 1/2\). Let \(q\) be the probability of getting a tail in a single toss. \(q = 1 - p = 1 - 1/2 = 1/2\).

The total number of possible outcomes when tossing 8 coins is \(2^8\).

Total outcomes = \(2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256\).

The probability of getting exactly \(k\) heads in \(n\) tosses is given by the binomial probability formula: \(P(X=k) = \binom{n}{k} p^k q^{n-k}\)

In this case, \(n=8\), \(p=1/2\), and \(q=1/2\). So, the formula becomes: \(P(X=k) = \binom{8}{k} (1/2)^k (1/2)^{8-k} = \binom{8}{k} (1/2)^{k + 8 - k} = \binom{8}{k} (1/2)^8 = \frac{\binom{8}{k}}{2^8} = \frac{\binom{8}{k}}{256}\)

Probability of Exactly 6 Heads

We calculate the probability of getting exactly 6 heads (\(k=6\)): \(P(X=6) = \binom{8}{6} (1/2)^8\) First, calculate the binomial coefficient \(\binom{8}{6}\): \(\binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8!}{6!2!} = \frac{8 \times 7 \times 6!}{6! \times 2 \times 1} = \frac{8 \times 7}{2} = 4 \times 7 = 28\) So, \(P(X=6) = 28 \times \frac{1}{256} = \frac{28}{256}\).

Probability of Exactly 7 Heads

Next, calculate the probability of getting exactly 7 heads (\(k=7\)): \(P(X=7) = \binom{8}{7} (1/2)^8\) Calculate the binomial coefficient \(\binom{8}{7}\): \(\binom{8}{7} = \frac{8!}{7!(8-7)!} = \frac{8!}{7!1!} = \frac{8 \times 7!}{7! \times 1} = 8\) So, \(P(X=7) = 8 \times \frac{1}{256} = \frac{8}{256}\).

Probability of Exactly 8 Heads

Finally, calculate the probability of getting exactly 8 heads (\(k=8\)): \(P(X=8) = \binom{8}{8} (1/2)^8\) Calculate the binomial coefficient \(\binom{8}{8}\): \(\binom{8}{8} = \frac{8!}{8!(8-8)!} = \frac{8!}{8!0!} = 1\) (since \(0! = 1\)) So, \(P(X=8) = 1 \times \frac{1}{256} = \frac{1}{256}\).

Probability of At Least 6 Heads

The probability of getting at least 6 heads is the sum of the probabilities of getting exactly 6, 7, or 8 heads: \(P(X \ge 6) = P(X=6) + P(X=7) + P(X=8)\) \(P(X \ge 6) = \frac{28}{256} + \frac{8}{256} + \frac{1}{256}\) \(P(X \ge 6) = \frac{28 + 8 + 1}{256} = \frac{37}{256}\)

So, the probability of getting at least 6 heads when 8 coins are tossed simultaneously is \(37/256\).

Revision Table: Coin Toss Probability

Additional Information: Binomial Distribution Basics

The binomial distribution is used for situations where:

• There is a fixed number of trials (\(n\)). In this case, 8 coin tosses.
• Each trial has only two possible outcomes (success or failure). Here, getting a head is a success, getting a tail is a failure.
• The probability of success (\(p\)) is constant for every trial. For a fair coin, \(p = 1/2\).
• The trials are independent. The outcome of one coin toss does not affect the outcome of another.

The formula \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) calculates the probability of observing exactly \(k\) successes in \(n\) trials. The term \(\binom{n}{k}\) (read as "n choose k") represents the number of different ways to choose \(k\) successes from \(n\) trials.