The problem asks for the probability of getting at least 5 heads when a fair coin is tossed 6 times. This scenario follows a binomial distribution.

Coin Toss Probability Calculation

We are tossing a fair coin 6 times ($n=6$). For a fair coin, the probability of getting a head ($p$) is 0.5, and the probability of getting a tail ($q$) is also 0.5.

The event "at least 5 heads" includes getting exactly 5 heads or exactly 6 heads.

Binomial Probability Formula

The formula for binomial probability is:

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

Where:

• $n$ = number of trials (tosses) = 6
• $k$ = number of successful outcomes (heads)
• $p$ = probability of success (head) = 0.5
• $q$ = probability of failure (tail) = 0.5

We need to calculate the probability for $k=5$ and $k=6$.

Probability of Exactly 5 Heads

Using the formula for $k=5$:

$ P(X=5) = \binom{6}{5} (0.5)^5 (0.5)^{6-5} $

$ P(X=5) = \binom{6}{5} (0.5)^5 (0.5)^1 $

$ P(X=5) = 6 \times (0.5)^6 $

$ P(X=5) = 6 \times \frac{1}{64} = \frac{6}{64} $

Probability of Exactly 6 Heads

Using the formula for $k=6$:

$ P(X=6) = \binom{6}{6} (0.5)^6 (0.5)^{6-6} $

$ P(X=6) = \binom{6}{6} (0.5)^6 (0.5)^0 $

$ P(X=6) = 1 \times (0.5)^6 \times 1 $

$ P(X=6) = 1 \times \frac{1}{64} = \frac{1}{64} $

Probability of At Least 5 Heads

To find the probability of at least 5 heads, we sum the probabilities of getting exactly 5 heads and exactly 6 heads:

$ P(X \ge 5) = P(X=5) + P(X=6) $

$ P(X \ge 5) = \frac{6}{64} + \frac{1}{64} $

$ P(X \ge 5) = \frac{7}{64} $

Result Conversion

Now, we convert the fraction $\frac{7}{64}$ to a decimal:

$ \frac{7}{64} \approx 0.109375 $

Rounding to two decimal places, the probability is 0.11.