Let a die be rolled n times. Let the probability of getting odd numbers seven times be equal to the

Q: Wait, why can we just cancel the $(1/2)^n$ term? Is that mathematically sound?

Absolutely! Since $(1/2)^n$ is a common factor on both sides of the equation and is strictly positive (it never equals zero), dividing both sides by it is a perfectly valid algebraic operation. It simplifies our life significantly!

Q: Why does $\binom{n}{7} = \binom{n}{9}$ imply $n=16$? I thought it could be something else.

Great question. The property $\binom{n}{x} = \binom{n}{y}$ implies either $x=y$ or $n = x+y$. Since $7 \neq 9$, the only remaining possibility is $n = 7+9 = 16$. It's a beautiful consequence of the symmetry in Pascal's Triangle.

Q: Why did we switch the definition of 'success' to 'even numbers' in the second part?

The binomial distribution is flexible! A 'success' is just the outcome we are interested in counting. Whether we call 'odd' a success or 'even' a success, the math remains consistent as long as we define our probability $p$ correctly for that specific event.

Question

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$, then $k$ is equal to

Options:

(A) 60

(B) 15

(C) 90

(D) 30

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a standard six-sided die and rolling it $n$ times. This scenario is a classic application of the **Binomial Distribution**.

We define a 'success' as rolling an odd number (1, 3, or 5). The probability $p$ of success on a single roll is:
$$p = \frac{3}{6} = \frac{1}{2}$$

Consequently, the probability $q$ of failure (rolling an even number) is:
$$q = 1 - p = \frac{1}{2}$$

The probability of obtaining exactly $r$ successes in $n$ trials is given by the formula:
$$P(X=r) = \binom{n}{r} p^r q^{n-r}$$

Substituting our specific probabilities, we obtain:
$$P(X=r) = \binom{n}{r} \left(\frac{1}{2}ight)^r \left(\frac{1}{2}ight)^{n-r} = \binom{n}{r} \left(\frac{1}{2}ight)^n$$

### The Hidden Symmetry

The problem states that the probability of getting exactly seven odd numbers is equal to the probability of getting exactly nine odd numbers. Mathematically, this is expressed as:
$$P(X=7) = P(X=9)$$

Substituting our derived formula into this equality, we get:
$$\binom{n}{7} \left(\frac{1}{2}ight)^n = \binom{n}{9} \left(\frac{1}{2}ight)^n$$

Since $\left(\frac{1}{2}ight)^n$ is non-zero, we can cancel it from both sides to arrive at:
$$\binom{n}{7} = \binom{n}{9}$$

Using the property of combinations where $\binom{n}{x} = \binom{n}{y}$ implies either $x=y$ or $n = x+y$, and noting that $7 
eq 9$, we conclude:
$$n = 7 + 9 = 16$$

### The Final Target

Now that we have determined the total number of trials $n=16$, we calculate the probability of getting exactly two even numbers. We redefine our success as 'getting an even number', where $p = \frac{1}{2}$ and $r = 2$.

Using the binomial formula:
$$P(Y=2) = \binom{16}{2} \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^{14} = \binom{16}{2} \left(\frac{1}{2}ight)^{16}$$

Calculating the combination $\binom{16}{2}$:
$$\binom{16}{2} = \frac{16 	imes 15}{2 	imes 1} = 120$$

Thus, the probability is:
$$P(Y=2) = \frac{120}{2^{16}}$$

To match the required form $\frac{k}{2^{15}}$, we simplify the expression:
$$\frac{120}{2 	imes 2^{15}} = \frac{60}{2^{15}}$$

By direct comparison, we find the final value:
**$k = 60$**

Animated Solution for Mathematics - Probability: Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2 ^{15}}$ , then $k$ is equal to

Select Answer:

Visualized Solution

Understanding the Experiment

The Binomial Formula

Setting up the Condition

Equating and Simplifying

Finding the Value of $n$

Target: Even Numbers Twice

Setting up the Target Probability

Atomic Computation of $(2 16)$

Matching the Form

Final Conclusion

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Hidden Symmetry

The Final Target

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