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Understanding complementary probability can simplify many probability problems. In the context of tossing a coin, complementary probability allows us to look at the problem from another angle, making it easier to solve certain types of questions.

When considering a fair coin toss, the probability of getting a head (denoted as \( P(H) \)) or a tail (denoted as \( P(T) \)) is both 0.5 since there are only two equally likely outcomes.

The complementary probability is particularly useful when we wish to calculate the probability of an event happening at least once, in our case, getting at least one head.

• The probability of not observing a head in one toss is the same as getting a tail, which is also 0.5.
• For multiple tosses, the probability of getting tails in every toss, known as the complementary probability, is \((0.5)^n\) where \(n\) is the number of tosses.

To find the probability of getting at least one head, you subtract the probability of the complementary event from 1:

\[ P(\text{at least one head}) = 1 - P(\text{all tails}) = 1 - (0.5)^n \]

Inequality is a powerful tool in determining how many times we need to repeat an experiment to achieve a certain outcome with a given probability. In this case, we look for the minimum number of coin tosses that ensure a probability of observing at least one head is at least 0.9.

We begin by setting up an inequality:

• The desired probability condition is \(1 - (0.5)^n \geq 0.9\).
• This translates to \((0.5)^n \leq 0.1\).

This inequality tells us how unlikely getting tails in every toss should be for us to have a greater than 0.9 probability of seeing at least one head.

By solving the inequality, we can determine the smallest number of coin tosses needed. It requires understanding the underlying mathematical expressions, which in this case, involves logarithms.

Logarithms serve as a bridge when dealing with exponential inequalities, which are common when working with probabilities repeated over multiple independent events. In our situation, we need to solve for \(n\) in the inequality \((0.5)^n \leq 0.1\).

Taking the logarithm of both sides can convert the exponential form into a linear form. Using base 2 logarithm (\(\log_2\)), we can express:

• \(n * \log_2(0.5) \leq \log_2(0.1)\).
• Knowing that \(\log_2(0.5) = -1\), we rearrange to solve for \(n\).

Thus, \(n \geq \frac{\log_2(0.1)}{-1}\).

Finally, calculate \(\log_2(0.1)\). Since \(0.1\) is \(1/10\), use the change of base formula or a calculator to find that \(n\approx 3.32\).
Since \(n\) must be a whole number, it is rounded up to ensure the probability condition holds, resulting in \(n = 4\) tosses. Logarithms simplify the process, making it easier to interpret and calculate such probability-related questions.