Key Concepts and Formulas

• Complementary Probability Rule: For any event $E$, the probability of $E$ occurring is $P(E) = 1 - P(E^c)$, where $E^c$ is the complement of event $E$ (the event that $E$ does not occur). This rule is particularly useful for "at least one" type of problems.
• Probability of Independent Events: If multiple events are independent (the outcome of one does not affect the others), the probability of all of them occurring is the product of their individual probabilities. For events $A_1, A_2, \dots, A_n$, $P(A_1 \text{ and } A_2 \text{ and } \dots \text{ and } A_n) = P(A_1) \times P(A_2) \times \dots \times P(A_n)$.
• Basic Coin Toss Probabilities: For a fair coin, the probability of getting a Head (H) is $P(H) = \frac{1}{2}$, and the probability of getting a Tail (T) is $P(T) = \frac{1}{2}$.

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Step-by-Step Solution

Step 1: Define the Event and its Complement

We are looking for the minimum number of tosses ($n$) such that the probability of observing "at least one head" is at least 90%.

• Let Event $E$ be "observing at least one head in $n$ tosses."
• The Complement of Event $E$ ($E^c$) is "observing no heads in $n$ tosses." This means all $n$ tosses must result in tails.

Using complementary probability simplifies the problem significantly because calculating $P(E^c)$ (all tails) is much easier than directly calculating $P(E)$ (summing probabilities for 1 head, 2 heads, ..., up to $n$ heads).

Step 2: Calculate the Probability of the Complement Event ($E^c$)

For a fair coin, the probability of getting a Tail in a single toss is $P(T) = \frac{1}{2}$. Since each coin toss is an independent event, the probability of getting tails in $n$ consecutive tosses is the product of the probabilities of getting a tail in each individual toss.

$P(E^c) = P(\text{Tail on 1st toss}) \times P(\text{Tail on 2nd toss}) \times \dots \times P(\text{Tail on } n\text{th toss})$ $P(E^c) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \dots \times \left(\frac{1}{2}\right) \quad (n \text{ times})$ $P(E^c) = \left(\frac{1}{2}\right)^n$

Step 3: Calculate the Probability of the Desired Event ($E$)

Now, we use the complementary probability rule: $P(E) = 1 - P(E^c)$ Substituting the expression for $P(E^c)$ from Step 2: $P(E) = 1 - \left(\frac{1}{2}\right)^n$

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

The problem states that the probability of observing at least one head must be "at least 90%." Converting 90% to a decimal, we get $0.9$. So, the condition is: $P(E) \ge 0.9$ Substituting the expression for $P(E)$ from Step 3: $1 - \left(\frac{1}{2}\right)^n \ge 0.9$

Step 5: Solve the Inequality for the Minimum Number of Tosses ($n$)

Our goal is to find the smallest integer value of $n$ that satisfies this inequality.

1. Isolate the term with $n$: Subtract 1 from both sides of the inequality. $-\left(\frac{1}{2}\right)^n \ge 0.9 - 1$ $-\left(\frac{1}{2}\right)^n \ge -0.1$

2. Multiply by -1 and reverse the inequality sign: To make the term with $n$ positive, multiply both sides by -1. Remember that multiplying or dividing an inequality by a negative number reverses the direction of the inequality sign. $\left(\frac{1}{2}\right)^n \le 0.1$

3. Rewrite and take reciprocals: Express $0.1$ as $\frac{1}{10}$ and $\left(\frac{1}{2}\right)^n$ as $\frac{1}{2^n}$. $\frac{1}{2^n} \le \frac{1}{10}$ Now, take the reciprocal of both sides. When taking the reciprocal of both sides of an inequality (where both sides are positive), the direction of the inequality sign must be reversed. $2^n \ge 10$

4. Find the minimum integer value of $n$: We need to find the smallest integer $n$ for which $2^n$ is greater than or equal to 10.

   • 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!)

   The minimum integer value of $n$ that satisfies the condition $2^n \ge 10$ is $n=4$.

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Common Mistakes & Tips

• Forgetting to Reverse Inequality Signs: This is a very common error. Always remember to reverse the inequality sign when multiplying/dividing by a negative number or when taking reciprocals of both sides (for positive values).
• Misinterpreting "At Least One": Directly calculating "at least one" head can be tedious. Always think of using the complementary probability rule ($1 - P(\text{no heads})$) for these types of problems.
• Checking Integer Solutions: After solving an inequality, especially when finding a minimum or maximum integer, it's good practice to test values around your calculated boundary to confirm the correct integer solution.

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Summary

To determine the minimum number of coin tosses required for the probability of observing at least one head to be at least 90%, we effectively used the complementary probability rule. By calculating the probability of the opposite event (no heads) as $(1/2)^n$, we set up the inequality $1 - (1/2)^n \ge 0.9$. Solving this inequality led to $2^n \ge 10$. By testing integer values, we found that the smallest integer $n$ satisfying this condition is 4.

The final answer is $\boxed{4}$, which corresponds to option (C).