Key Concepts and Formulas

• Complementary Probability: The probability of an event $E$ occurring is $P(E) = 1 - P(E')$, where $E'$ is the complement of $E$ (the event that $E$ does not occur). This is an incredibly powerful tool for problems involving "at least one" scenarios, as calculating the complement is often much simpler.
• Probability of Independent Events: If multiple events are independent (the outcome of one does not affect the others), the probability that all of them occur is the product of their individual probabilities. For example, the outcome of each coin toss is independent of previous tosses.
• Solving Exponential Inequalities: To find the value of a variable in an exponent, logarithms are typically employed. It is crucial to remember to reverse the inequality sign if you multiply or divide both sides by a negative number.

---

Step-by-Step Solution

Step 1: Define Probabilities for a Single Toss of a Fair Coin

• What we do: Establish the fundamental probabilities for the outcomes of a single coin toss.
• Why we do it: These are the basic probabilities that will be used to calculate probabilities over multiple tosses. For a fair coin, there are two equally likely outcomes: Head (H) and Tail (T). The probability of getting a Head in a single toss is: $P(H) = \frac{1}{2}$ The probability of getting a Tail in a single toss is: $P(T) = \frac{1}{2}$
• Reasoning: A "fair coin" implies that each outcome has an equal chance of occurring. Since there are two possible outcomes, each must have a probability of $1/2$.

Step 2: Calculate the Probability of the Complementary Event

• What we do: Define the event we are interested in, identify its complement, and then calculate the probability of this simpler complementary event.
• Why we do it: The event "getting at least one head" is complex to calculate directly (it would involve summing probabilities of 1 head, 2 heads, ..., up to $n$ heads). Its complement, "getting no heads (i.e., all tails)," is significantly simpler to calculate directly, leveraging the independence of coin tosses. Let $E$ be the event "getting at least one head in $n$ tosses." The complementary event, $E'$, is "getting no heads in $n$ tosses." This means that all $n$ tosses must result in tails. Since each coin toss is an independent event (the outcome of one toss does not influence subsequent tosses), the probability of a sequence of specific outcomes is the product of their individual probabilities. So, the probability of getting $n$ tails consecutively is: $P(E') = P(\text{all tails}) = P(T_1 \cap T_2 \cap \dots \cap T_n)$ $P(E') = P(T_1) \times P(T_2) \times \dots \times P(T_n)$ Substituting $P(T) = 1/2$ from Step 1: $P(E') = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \dots \times \left(\frac{1}{2}\right) \quad (n \text{ times})$ $P(E') = \left(\frac{1}{2}\right)^n$

Step 3: Apply the Complementary Probability Formula

• What we do: Use the complementary probability principle to find $P(E)$.
• Why we do it: This step efficiently connects our easily calculated complementary probability back to the original, more complex event we are interested in. Using the formula $P(E) = 1 - P(E')$: $P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^n$

Step 4: Set Up the Inequality Based on the Given Condition

• What we do: Translate the problem's requirement into a mathematical inequality.
• Why we do it: The problem specifies a minimum probability threshold that our calculated probability must meet, which needs to be expressed mathematically to solve for $n$. The problem states that the probability of getting at least one head must be "at least 0.9." This means it must be greater than or equal to 0.9. $P(\text{at least one head}) \ge 0.9$ Substitute the expression for $P(\text{at least one head})$ from Step 3: $1 - \left(\frac{1}{2}\right)^n \ge 0.9$

Step 5: Solve the Inequality for the Exponential Term

• What we do: Algebraically manipulate the inequality to isolate the term containing $n$.
• Why we do it: Isolating the exponential term $\left(\frac{1}{2}\right)^n$ makes it easier to determine the possible values of $n$. First, subtract 1 from both sides of the inequality: $-\left(\frac{1}{2}\right)^n \ge 0.9 - 1$ $-\left(\frac{1}{2}\right)^n \ge -0.1$ Next, multiply both sides by -1. Crucially, remember to reverse the inequality sign when multiplying or dividing by a negative number: $\left(\frac{1}{2}\right)^n \le 0.1$ To work with fractions, we can convert $0.1$ to $\frac{1}{10}$: $\left(\frac{1}{2}\right)^n \le \frac{1}{10}$

Step 6: Determine the Minimum Integer Value of $n$ (Trial and Error or Logarithms)

• What we do: Test integer values for $n$ or use logarithms to find the smallest integer that satisfies the inequality.
• Why we do it: $n$ represents the number of tosses, which must be a positive integer. We are looking for the minimum such integer.

We need to find the smallest positive integer $n$ that satisfies $\left(\frac{1}{2}\right)^n \le \frac{1}{10}$. This inequality can be rewritten as $2^n \ge 10$. Let's test values for $n$ systematically, starting from $n=1$:

• For $n=1$: $2^1 = 2$. Is $2 \ge 10$? No.
• For $n=2$: $2^2 = 4$. Is $4 \ge 10$? No.
• For $n=3$: $2^3 = 8$. Is $8 \ge 10$? No.
• For $n=4$: $2^4 = 16$. Is $16 \ge 10$? Yes.

Since $n$ must be an integer (number of tosses), the first integer value that satisfies the condition is $n=4$.

Alternative Method: Using Logarithms $\left(\frac{1}{2}\right)^n \le \frac{1}{10}$ Taking $\log_2$ of both sides: $\log_2\left(\left(\frac{1}{2}\right)^n\right) \le \log_2\left(\frac{1}{10}\right)$ Using logarithm properties $\log(a^b) = b \log a$ and $\log(1/x) = -\log x$: $n \log_2\left(\frac{1}{2}\right) \le \log_2(1) - \log_2(10)$ $n(-1) \le 0 - \log_2(10)$ $-n \le -\log_2(10)$ Multiply by -1 and reverse the inequality sign: $n \ge \log_2(10)$ Since $2^3 = 8$ and $2^4 = 16$, we know that $\log_2(10)$ lies between 3 and 4. Approximately, $\log_2(10) \approx 3.32$. Thus, $n \ge 3.32...$. As $n$ must be an integer, the smallest integer value for $n$ that satisfies this condition is $n=4$.

---

Common Mistakes & Tips

• Inequality Sign Reversal: Always remember to reverse the inequality sign when multiplying or dividing both sides by a negative number. This is a common source of error.
• "At Least One" Shortcut: For problems involving "at least one" of an event, immediately consider using complementary probability ($P(E) = 1 - P(E')$) as it almost always simplifies the calculation.
• Integer Constraints: Physical quantities like the number of tosses must be integers. After solving an inequality, ensure you select the appropriate integer value (minimum or maximum) based on the problem's context.

---

Summary This problem effectively demonstrates the power of complementary probability in simplifying complex "at least one" scenarios. We calculated the probability of the complementary event, "getting no heads" (all tails), as $(1/2)^n$. Subtracting this from 1 gives the probability of "at least one head," which is $1 - (1/2)^n$. We then set this probability to be at least 0.9, leading to the inequality $1 - (1/2)^n \ge 0.9$. Solving this inequality, either by testing integer values or using logarithms, reveals that $n$ must be greater than or equal to approximately 3.32. Since $n$ must be a whole number of tosses, the minimum integer value of $n$ that satisfies the condition is 4.

The final answer is \boxed{4}.