suppose that a fair coin is tossed independently n times. the probability of obtaining exactly n-1 heads, given that at least n-2 heads are obtained is

We are given that a fair coin is tossed independently \( n \) times, and we need to find the conditional probability of obtaining exactly \( n-1 \) heads, given that at least \( n-2 \) heads are obtained.

### Step 1: Define events
- Let \( A \) be the event that exactly \( n-1 \) heads are obtained.
- Let \( B \) be the event that at least \( n-2 \) heads are obtained. This includes the cases of getting \( n-2 \), \( n-1 \), or \( n \) heads.

We are asked to compute \( P(A \mid B) \), the conditional probability of \( A \) given \( B \). By the definition of conditional probability, this is:

\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
\]

### Step 2: Compute the probability of \( A \cap B \)
The event \( A \cap B \) is simply the event \( A \), because if you have exactly \( n-1 \) heads, then you automatically satisfy the condition of having at least \( n-2 \) heads. Therefore:

\[
P(A \cap B) = P(A)
\]

Since the probability of getting exactly \( k \) heads in \( n \) coin tosses follows a binomial distribution, we have:

\[
P(A) = P(\text{exactly } n-1 \text{ heads}) = \binom{n}{n-1} \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^n
\]

### Step 3: Compute the probability of \( B \)
The event \( B \) includes getting \( n-2 \), \( n-1 \), or \( n \) heads. Thus:

\[
P(B) = P(\text{exactly } n-2 \text{ heads}) + P(\text{exactly } n-1 \text{ heads}) + P(\text{exactly } n \text{ heads})
\]

These are all binomial probabilities:

\[
P(B) = \binom{n}{n-2} \left(\frac{1}{2}\right)^n + \binom{n}{n-1} \left(\frac{1}{2}\right)^n + \binom{n}{n} \left(\frac{1}{2}\right)^n
\]

Simplifying the binomial coefficients:

\[
P(B) = \binom{n}{2} \left(\frac{1}{2}\right)^n + n \left(\frac{1}{2}\right)^n + 1 \cdot \left(\frac{1}{2}\right)^n
\]

Factoring out \( \left(\frac{1}{2}\right)^n \):

\[
P(B) = \left(\frac{1}{2}\right)^n \left[\binom{n}{2} + n + 1\right]
\]

### Step 4: Compute the conditional probability
Now, using the formula for conditional probability:

\[
P(A \mid B) = \frac{P(A)}{P(B)} = \frac{n \left(\frac{1}{2}\right)^n}{\left(\frac{1}{2}\right)^n \left[\binom{n}{2} + n + 1\right]}
\]

The \( \left(\frac{1}{2}\right)^n \) terms cancel out, leaving:

\[
P(A \mid B) = \frac{n}{\binom{n}{2} + n + 1}
\]

### Final Answer:
\[
P(A \mid B) = \frac{n}{\frac{n(n-1)}{2} + n + 1} = \frac{n}{\frac{n(n-1)}{2} + n + 1}
\]

This is the conditional probability of obtaining exactly \( n-1 \) heads, given that at least \( n-2 \) heads are obtained.

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Would you like further clarification or a detailed breakdown of any part of the solution?

Here are 5 related questions for practice:
1. What is the probability of obtaining exactly \( n \) heads in \( n \) tosses?
2. How does the probability change if the coin is biased instead of fair?
3. What is the probability of obtaining exactly \( n-2 \) heads in \( n \) tosses?
4. How would you calculate the expected number of heads in \( n \) coin tosses?
5. What is the probability of obtaining at least one head in \( n \) tosses?

**Tip**: Conditional probabilities often require careful attention to the definition of events. Always ensure that the "given" event includes all relevant outcomes.

Learn how to calculate the conditional probability of obtaining exactly n-1 heads given that at least n-2 heads are obtained in n independent coin tosses. This problem explores key probability concepts, including binomial distribution and conditional probability, suitable for advanced high school and college students.

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