Understanding the Coin Toss Probability Problem

This problem asks us to determine the expected number of times we would observe "at least three heads" when tossing seven unbiased coins simultaneously, repeating this experiment 128 times. This scenario fits the description of a binomial probability distribution.

Let's define the parameters for a single throw of seven unbiased coins:

• Number of trials (coins), \(n = 7\)
• Probability of success (getting a head) on a single trial, \(p = 0.5\) (since the coin is unbiased)
• Probability of failure (getting a tail) on a single trial, \(q = 1 - p = 0.5\)

The number of heads obtained in a single throw of 7 coins follows a binomial distribution, denoted as \(X \sim B(n=7, p=0.5)\). The probability of getting exactly \(k\) heads in 7 tosses is given by the binomial probability formula:

\(P(X=k) = \binom{n}{k} p^k q^{n-k}\)

Substituting the values for our problem (\(n=7, p=0.5, q=0.5\)):

\(P(X=k) = \binom{7}{k} (0.5)^k (0.5)^{7-k} = \binom{7}{k} (0.5)^7\)

Calculating Probability of At Least Three Heads

We are interested in the number of throws where we find "at least three heads". This means the number of heads, \(k\), is 3 or more, i.e., \(k \ge 3\). The possible values for \(k\) are 3, 4, 5, 6, or 7.

Calculating \(P(X \ge 3)\) directly would involve summing the probabilities \(P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)\). An easier approach is to calculate the probability of the complementary event, which is getting "less than three heads" (\(k < 3\)), and subtract it from 1. The event \(k < 3\) corresponds to getting 0, 1, or 2 heads.

\(P(X \ge 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]\)

Calculating Probabilities for 0, 1, and 2 Heads

We need to calculate \(P(X=k) = \binom{7}{k} (0.5)^7\) for \(k=0, 1, 2\). Note that \((0.5)^7 = \left(\frac{1}{2}\right)^7 = \frac{1}{128}\).

• For \(k=0\): \(P(X=0) = \binom{7}{0} (0.5)^7 = 1 \times \frac{1}{128} = \frac{1}{128}\)
• For \(k=1\): \(P(X=1) = \binom{7}{1} (0.5)^7 = 7 \times \frac{1}{128} = \frac{7}{128}\)
• For \(k=2\): \(P(X=2) = \binom{7}{2} (0.5)^7 = \frac{7 \times 6}{2 \times 1} \times \frac{1}{128} = 21 \times \frac{1}{128} = \frac{21}{128}\)

Summing Probabilities for Less Than Three Heads

Now, sum the probabilities for \(k=0, 1, 2\):

\(P(X < 3) = P(X=0) + P(X=1) + P(X=2) = \frac{1}{128} + \frac{7}{128} + \frac{21}{128} = \frac{1 + 7 + 21}{128} = \frac{29}{128}\)

Calculating Probability of At Least Three Heads

Using the complement rule:

\(P(X \ge 3) = 1 - P(X < 3) = 1 - \frac{29}{128} = \frac{128 - 29}{128} = \frac{99}{128}\)

This is the probability of getting at least three heads in a single throw of seven unbiased coins.

Expected Number of Throws with At Least Three Heads

The experiment of tossing seven coins is repeated 128 times. The expected number of times an event occurs is the probability of the event occurring in a single trial multiplied by the total number of trials.

• Total number of throws, \(N = 128\)
• Probability of getting at least three heads in one throw, \(P(X \ge 3) = \frac{99}{128}\)

Expected number of throws with at least three heads = \(N \times P(X \ge 3)\)

Expected Number = \(128 \times \frac{99}{128} = 99\)

Therefore, in 128 throws of seven unbiased coins, we would expect to find at least three heads in 99 throws.

From the table, the sum of probabilities for \(k \ge 3\) is \(P(X \ge 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = \frac{35}{128} + \frac{35}{128} + \frac{21}{128} + \frac{7}{128} + \frac{1}{128} = \frac{35+35+21+7+1}{128} = \frac{99}{128}\). This confirms the result obtained using the complement rule.

Revision Table: Key Concepts

Additional Information: Binomial Coefficients

The term \(\binom{n}{k}\) in the binomial probability formula is a binomial coefficient, read as "n choose k". It represents the number of ways to choose \(k\) items from a set of \(n\) items without regard to the order of selection. It is calculated using the formula:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

where \(n!\) (n factorial) is the product of all positive integers up to \(n\), with \(0! = 1\).

For example, \(\binom{7}{2}\) is the number of ways to get exactly 2 heads in 7 tosses.

\(\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1)} = \frac{7 \times 6}{2 \times 1} = 21\)

These binomial coefficients are crucial for calculating the probabilities in a binomial distribution problem like this coin toss scenario.