An ordinary dice is rolled for a certain number of times. If the probability of getting an odd numbe

Q: Why can we just cancel $(\frac{1}{2})^n$ from both sides? Isn't that dangerous?

Great question! In algebra, we can cancel terms as long as they are non-zero. Since $(\frac{1}{2})^n$ is strictly positive for any finite $n$, it is perfectly safe to divide both sides by it. It is just a scaling factor that doesn't affect the equality of the combinations.

Q: How did you get $n=5$ so quickly from $\binom{n}{2} = \binom{n}{3}$?

Ah, the beauty of Pascal's Triangle! Remember the identity $\binom{n}{r} = \binom{n}{n-r}$. If $\binom{n}{x} = \binom{n}{y}$ and $x \neq y$, then $x + y$ must equal $n$. Here, $2 + 3 = 5$. It is a fundamental property that saves you from expanding factorials!

Q: Is there a shortcut for the final sum?

Yes! When $p=q=1/2$, the distribution is perfectly symmetric. The sum of probabilities for odd terms is always $1/2$. It is a beautiful result of the Binomial Theorem expansion $(p+q)^n - (q-p)^n$.

Question

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

Options:

(A) $\frac{3}{16}$

(B) $\frac{1}{2}$

(C) $\frac{5}{16}$

(D) $\frac{1}{32}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at a table, holding an ordinary six-sided die. You are about to roll it multiple times. Each roll is a fresh start, an independent event, a tiny universe of chance. This is the essence of the Binomial Distribution.

First, we must define our single trial. When you roll a die, the probability of getting an odd number—$1, 3,$ or $5$—is $p = \frac{3}{6} = \frac{1}{2}$.

Consequently, the probability of getting an even number is $q = 1 - p = \frac{1}{2}$. Since we are rolling the die $n$ times, we are looking at a sequence of Bernoulli trials where the probability of success and failure are perfectly balanced.

### The Algebraic Elegance of Symmetry

The problem presents us with a fascinating condition: the probability of getting an odd number exactly twice is equal to the probability of getting an even number exactly three times. Let $X$ be the random variable representing the number of odd outcomes.

We are given $P(X=2) = P(X=3)$. Using the binomial formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$, we write:

$$ \binom{n}{2} \left( \frac{1}{2} ight)^2 \left( \frac{1}{2} ight)^{n-2} = \binom{n}{3} \left( \frac{1}{2} ight)^3 \left( \frac{1}{2} ight)^{n-3} $$

Notice the magic here? Both sides simplify to the following expression:

$$ \binom{n}{2} \left( \frac{1}{2} ight)^n = \binom{n}{3} \left( \frac{1}{2} ight)^n $$

Because $\left( \frac{1}{2} ight)^n$ is never zero, we can divide it away, leaving us with the elegant equality $\binom{n}{2} = \binom{n}{3}$.

This is where your intuition should kick in. The property $\binom{n}{x} = \binom{n}{y}$ implies that either $x=y$ (which is impossible here, as $2 
eq 3$) or $x+y=n$.

Thus, $n = 2 + 3 = 5$. We have cracked the code: the die is rolled exactly five times.

### The Final Summation

Now, we need the probability of getting an odd number for an odd number of times. With $n=5$, this means we need the probability of getting $1, 3,$ or $5$ odd numbers.

We calculate the sum:

$$ P(X \in \{1, 3, 5\}) = \binom{5}{1}\left(\frac{1}{2}ight)^5 + \binom{5}{3}\left(\frac{1}{2}ight)^5 + \binom{5}{5}\left(\frac{1}{2}ight)^5 $$

Factoring out $\left(\frac{1}{2}ight)^5$, we get:

$$ \frac{1}{32} \left[ \binom{5}{1} + \binom{5}{3} + \binom{5}{5} ight] $$

Evaluating the combinations, we find $\binom{5}{1} = 5$, $\binom{5}{3} = 10$, and $\binom{5}{5} = 1$. The sum is $5 + 10 + 1 = 16$.

Finally, we calculate the result:

$$ \frac{16}{32} = \frac{1}{2} $$

This result is not a coincidence. Whenever $p=q=1/2$, the binomial distribution is perfectly symmetric, and the sum of probabilities for odd successes will always be exactly **1/2**. You have just mastered a fundamental principle of probability!

Animated Solution for Mathematics - Probability: An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

Select Answer:

Visualized Solution

Define Single Trial Probabilities

Identify the Distribution

Set Up the Given Condition

Equate and Simplify

Solve for $n$

Define the Target Probability

Substitute and Calculate

Final Summation

Conclusion and Takeaway

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Algebraic Elegance of Symmetry

The Final Summation

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