Key Concepts and Formulas

• Binomial Distribution: A random variable $X$ follows a Binomial distribution, denoted as $B(n, p)$, if it represents the number of successes in $n$ independent Bernoulli trials.
  • $n$: Total number of trials.
  • $p$: Probability of success in a single trial.
  • $q$: Probability of failure in a single trial, where $q = 1-p$.
• Probability Mass Function (PMF): 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 ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.
• Properties of Binomial Coefficients:
  • ${}^n C_0 = 1$
  • ${}^n C_1 = n$
  • ${}^n C_r = {}^n C_{n-r}$ (e.g., ${}^5 C_2 = {}^5 C_3$)

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Step-by-Step Solution

Given that the random variable $X$ follows a Binomial distribution $B(5, p)$, we know that $n=5$. Our goal is to first determine the unknown probability $p$ using the given condition $P(X=0) = P(X=1)$, and then use these parameters to calculate the required ratio $\frac{P(X=2)}{P(X=3)}$.

Step 1: Determine the Probability of Success ($p$) and Failure ($q$)

We are given the condition $P(X=0) = P(X=1)$. We will use the PMF to express these probabilities in terms of $p$ and $q$ and then solve for them.

1. Calculate $P(X=0)$: Using the PMF with $n=5$ and $r=0$: $P(X = 0) = {}^5 C_0 p^0 q^{5-0}$ Since ${}^5 C_0 = 1$ and $p^0 = 1$ (assuming $p \ne 0$), this simplifies to: $P(X = 0) = 1 \cdot 1 \cdot q^5 = q^5$

2. Calculate $P(X=1)$: Using the PMF with $n=5$ and $r=1$: $P(X = 1) = {}^5 C_1 p^1 q^{5-1}$ Since ${}^5 C_1 = 5$ and $p^1 = p$, this simplifies to: $P(X = 1) = 5pq^4$

3. Equate $P(X=0)$ and $P(X=1)$ and solve for $p$: From the given condition: $q^5 = 5pq^4$ To solve for $p$, we can divide both sides by $q^4$. We must ensure $q \ne 0$.

   • If $q=0$, then $p=1$ (since $p+q=1$). In this case, $P(X=0) = 0$ and $P(X=1) = 0$, so the condition $P(X=0)=P(X=1)$ would be satisfied ($0=0$).
   • However, if $p=1$ and $q=0$, then $P(X=2) = {}^5 C_2 (1)^2 (0)^3 = 0$ and $P(X=3) = {}^5 C_3 (1)^3 (0)^2 = 0$. The ratio $\frac{P(X=2)}{P(X=3)}$ would be $\frac{0}{0}$, which is undefined. Since the problem asks for a specific numerical value, we can conclude that $q \ne 0$. Thus, we can safely divide by $q^4$: $\frac{q^5}{q^4} = 5p$ $q = 5p$ Now, substitute $q = 1-p$ (from $p+q=1$): $1 - p = 5p$ $1 = 6p$ $p = \frac{1}{6}$ Finally, calculate $q$: $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$ So, the parameters of our Binomial distribution are $n=5$, $p=\frac{1}{6}$, and $q=\frac{5}{6}$.

Step 2: Calculate the Required Ratio $\frac{P(X=2)}{P(X=3)}$

Now that we have $p$ and $q$, we can calculate the ratio.

1. Express $P(X=2)$ and $P(X=3)$ using the PMF: For $X=2$: $P(X = 2) = {}^5 C_2 p^2 q^{5-2} = {}^5 C_2 p^2 q^3$ For $X=3$: $P(X = 3) = {}^5 C_3 p^3 q^{5-3} = {}^5 C_3 p^3 q^2$

2. Form the ratio and simplify: $\frac{P(X = 2)}{P(X = 3)} = \frac{{}^5 C_2 p^2 q^3}{{}^5 C_3 p^3 q^2}$ First, let's calculate the binomial coefficients:

   • ${}^5 C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$
   • ${}^5 C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$ (Note: ${}^5 C_2 = {}^5 C_{5-2} = {}^5 C_3$, so their values are equal.)

   Substitute these values into the ratio and simplify the powers of $p$ and $q$: $\frac{P(X = 2)}{P(X = 3)} = \frac{10 \cdot p^2 \cdot q^3}{10 \cdot p^3 \cdot q^2}$ The $10$s cancel. Using exponent rules, $\frac{q^3}{q^2} = q$ and $\frac{p^2}{p^3} = \frac{1}{p}$. $\frac{P(X = 2)}{P(X = 3)} = \frac{q}{p}$

3. Substitute the values of $p$ and $q$: We found $p = \frac{1}{6}$ and $q = \frac{5}{6}$. $\frac{P(X = 2)}{P(X = 3)} = \frac{5/6}{1/6}$ $\frac{P(X = 2)}{P(X = 3)} = \frac{5}{6} \times \frac{6}{1} = 5$

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Common Mistakes & Tips

• Forgetting $p+q=1$: This fundamental relationship is essential for solving for $p$ and $q$ when only one equation is given.
• Errors in Binomial Coefficient Calculation: Be careful when calculating ${}^n C_r$. Utilizing properties like ${}^n C_r = {}^n C_{n-r}$ can save time and prevent errors.
• Algebraic Simplification: When dealing with ratios of probabilities, terms often simplify significantly. Carefully apply exponent rules and cancel common factors.
• Checking for Division by Zero: Always consider if a variable you are dividing by (like $q^4$ in Step 1) could be zero. Analyze what that case would imply for the problem to determine if it's a valid scenario or can be ruled out.
• General Ratio Formula: For competitive exams, it's useful to know the general ratio formula for consecutive Binomial probabilities: $\frac{P(X=r+1)}{P(X=r)} = \frac{n-r}{r+1} \cdot \frac{p}{q}$ The inverse, $\frac{P(X=r)}{P(X=r+1)} = \frac{r+1}{n-r} \cdot \frac{q}{p}$, can be directly applied here with $n=5, r=2$: $\frac{P(X=2)}{P(X=3)} = \frac{2+1}{5-2} \cdot \frac{q}{p} = \frac{3}{3} \cdot \frac{q}{p} = \frac{q}{p}$ This confirms our simplification and provides a quicker way to arrive at $\frac{q}{p}$.

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Summary

This problem effectively demonstrates a two-step approach to solving Binomial distribution questions. First, we utilized the given condition $P(X=0) = P(X=1)$ along with the Binomial PMF and the relation $p+q=1$ to determine the specific probabilities of success ($p=\frac{1}{6}$) and failure ($q=\frac{5}{6}$). Second, we applied these parameters to calculate the required ratio $\frac{P(X=2)}{P(X=3)}$. Careful algebraic simplification, particularly of the binomial coefficients and powers of $p$ and $q$, led to the final numerical answer.

The final answer is $\boxed{5}$, which corresponds to option (D).