1. Key Concepts and Formulas

• Binomial Probability Distribution: This distribution models the probability of a certain number of "successes" in a fixed number of independent trials, where each trial has only two possible outcomes (success or failure). The probability of getting exactly $r$ successes in $n$ trials is given by: $P(X=r) = {^n C_r} p^r q^{n-r}$ where:
  • $n$ is the total number of trials.
  • $r$ is the number of successes.
  • $p$ is the probability of success in a single trial.
  • $q = 1-p$ is the probability of failure in a single trial.
  • ${^n C_r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.
• Properties of Binomial Coefficients: A crucial property for solving problems involving equal probabilities is that if ${^n C_a} = {^n C_b}$, then either $a = b$ or $a+b = n$. This property is derived from the symmetry of combinations, where choosing $a$ items is the same as choosing to leave out $n-a$ items, i.e., ${^n C_a} = {^n C_{n-a}}$.
• Probability for a Fair Die: For a standard 6-sided die, there are 3 odd numbers ({1, 3, 5}) and 3 even numbers ({2, 4, 6}). Thus, the probability of rolling an odd number is $3/6 = 1/2$, and the probability of rolling an even number is $3/6 = 1/2$.

2. Step-by-Step Solution

Step 1: Determine Basic Probabilities for a Single Die Roll

• What we are doing: Establishing the probabilities of getting an odd or an even number in a single roll of a fair die.
• Why: These fundamental probabilities ($p$ and $q$) are essential inputs for the binomial probability formula in subsequent steps.
• A standard fair die has 6 faces with outcomes $\{1, 2, 3, 4, 5, 6\}$.
• Number of odd outcomes: $\{1, 3, 5\}$, which is 3 outcomes.
• Number of even outcomes: $\{2, 4, 6\}$, which is 3 outcomes.
• The probability of getting an odd number ($p_{\text{odd}}$) in a single roll is: $p_{\text{odd}} = \frac{\text{Number of odd outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$
• The probability of getting an even number ($p_{\text{even}}$) in a single roll is: $p_{\text{even}} = \frac{\text{Number of even outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$
• Since both $p_{\text{odd}}$ and $p_{\text{even}}$ are $1/2$, for any event (getting an odd or an even number), the probability of 'success' ($p$) will be $1/2$, and the probability of 'failure' ($q = 1-p$) will also be $1/2$. This significantly simplifies calculations.

Step 2: Use the Given Equality to Find the Total Number of Rolls ($n$)

• What we are doing: Utilizing the first piece of information provided in the question to determine $n$, the total number of times the die was rolled.
• Why: The binomial probability formula requires the value of $n$. The equality of two probabilities allows us to set up an equation and solve for $n$.
• The problem states: "the probability of getting odd numbers seven times is equal to the probability of getting odd numbers nine times."
• In this context, 'success' is getting an odd number, so $p = p_{\text{odd}} = 1/2$, and $q = 1-p = 1/2$.
• Using the binomial probability formula $P(X=r) = {^n C_r} p^r q^{n-r}$:
  • Probability of getting odd numbers 7 times ($r=7$): $P(X=7) = {^n C_7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{n-7} = {^n C_7} \left(\frac{1}{2}\right)^{7+(n-7)} = {^n C_7} \left(\frac{1}{2}\right)^n$
  • Probability of getting odd numbers 9 times ($r=9$): $P(X=9) = {^n C_9} \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{n-9} = {^n C_9} \left(\frac{1}{2}\right)^{9+(n-9)} = {^n C_9} \left(\frac{1}{2}\right)^n$
• Equating these two probabilities as given by the problem: ${^n C_7} \left(\frac{1}{2}\right)^n = {^n C_9} \left(\frac{1}{2}\right)^n$
• Since $\left(\frac{1}{2}\right)^n$ is a common factor on both sides and is non-zero (as $n$ must be a positive integer), we can cancel it out: ${^n C_7} = {^n C_9}$
• Now, we apply the property of binomial coefficients: if ${^n C_a} = {^n C_b}$, then either $a = b$ or $a+b = n$.
• Clearly, $7 \neq 9$. Therefore, the second condition must hold: $n = 7 + 9$ $n = 16$
• So, the die was rolled a total of 16 times.

Step 3: Calculate the Probability of Getting Even Numbers Twice

• What we are doing: Using the determined value of $n$ to calculate the specific probability required by the second part of the question.
• Why: This calculation will yield an expression that can be equated to $\frac{k}{2^{15}}$ to find $k$.
• For this part, 'success' is getting an even number. From Step 1, $p' = p_{\text{even}} = 1/2$, and $q' = 1-p' = 1/2$.
• We want to find the probability of getting $r=2$ even numbers in $n=16$ trials.
• Using the binomial probability formula: $P(\text{even numbers twice}) = {^{16} C_2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{16-2}$ $P(\text{even numbers twice}) = {^{16} C_2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{14}$
• Combine the powers of $\frac{1}{2}$: $P(\text{even numbers twice}) = {^{16} C_2} \left(\frac{1}{2}\right)^{2+14} = {^{16} C_2} \left(\frac{1}{2}\right)^{16}$
• Now, calculate the binomial coefficient ${^{16} C_2}$: ${^{16} C_2} = \frac{16!}{2!(16-2)!} = \frac{16!}{2!14!} = \frac{16 \times 15 \times 14!}{2 \times 1 \times 14!} = \frac{16 \times 15}{2} = 8 \times 15 = 120$
• Substitute this value back into the probability expression: $P(\text{even numbers twice}) = 120 \times \left(\frac{1}{2}\right)^{16} = \frac{120}{2^{16}}$

Step 4: Equate and Solve for $k$

• What we are doing: Setting our calculated probability equal to the expression given in the problem to solve for the unknown $k$.
• Why: This is the final step to determine the value of $k$ as requested by the question.
• The problem states that the probability of getting even numbers twice is $\frac{k}{2^{15}}$.
• We have calculated this probability to be $\frac{120}{2^{16}}$.
• Equating the two expressions: $\frac{k}{2^{15}} = \frac{120}{2^{16}}$
• To solve for $k$, multiply both sides of the equation by $2^{15}$: $k = \frac{120}{2^{16}} \times 2^{15}$
• To simplify, recall that $2^{16}$ can be written as $2 \times 2^{15}$: $k = \frac{120}{2 \times 2^{15}} \times 2^{15}$
• The $2^{15}$ terms in the numerator and denominator cancel out: $k = \frac{120}{2}$ $k = 60$

3. Common Mistakes & Tips

• Misinterpreting the Binomial Coefficient Property: A common error is to assume ${^n C_a} = {^n C_b}$ only implies $a=b$. Remember the symmetric property $a+b=n$ is equally important.
• Errors in Exponent Simplification: When dealing with powers of $1/2$, it's crucial to correctly combine exponents (e.g., $(1/2)^x (1/2)^y = (1/2)^{x+y}$) and simplify fractions like $\frac{2^A}{2^B} = 2^{A-B}$.
• Calculation Mistakes: Be careful when computing combinations like ${^{16} C_2}$ and performing the final division.

4. Summary This problem is a comprehensive application of the Binomial Probability Distribution. We began by establishing the basic probabilities for odd and even numbers from a fair die. Next, we used the given equality of probabilities for 7 and 9 odd numbers to determine the total number of rolls, $n=16$, leveraging a key property of binomial coefficients. Finally, we calculated the probability of getting even numbers twice using $n=16$ and equated this to the given expression $\frac{k}{2^{15}}$ to solve for $k$. The result of our calculation for $k$ is 60.

5. Final Answer The final answer is $\boxed{\text{60}}$, which corresponds to option (B).