Problem 108 - A Complete Resource Book in Mathematics for JEE Main

Question

\(X\) folows a binomial distribution with parameters \(n\) and \(p\), and \(Y\) follows a binomial with parameters \(m\) and \(p\). Then, if \(X\) and \(Y\) are independent, \(P(X=r \mid X+Y=r+s)=\) (A) \(\left({ }^{\mathrm{m}} C_{\mathrm{r}}\right)\left({ }^{\mathrm{n}} C_{\mathrm{s}}\right)^{/ \mathrm{m}+\mathrm{n}} \mathrm{C}_{\mathrm{r}+\mathrm{s}}\) (B) \(3^{\mathrm{m}} \mathrm{C}_{\mathrm{r}+\mathrm{s}} / \mathrm{m}+\mathrm{c} \mathrm{C}_{\mathrm{r}+\mathrm{s}}\) (C) \(2\left({ }^{\mathrm{m}} C_{\mathrm{r}}\right)\left({ }^{3} \mathrm{C}_{s}\right)^{/ \mathrm{m}+{ }^{\mathrm{r}}} \mathrm{C}_{\mathrm{r}+\mathrm{s}}\) (D) none of these

Step-by-Step Solution

Answer

(D) none of these

Step 1: Identify the Program

We need to find \( P(X = r \mid X + Y = r + s) \). This is a conditional probability problem involving two binomial random variables \( X \) and \( Y \).

Step 2: Utilize the Law of Total Probability

According to the law of conditional probability, \( P(X = r \mid X + Y = r + s) = \frac{P(X = r, X + Y = r + s)}{P(X + Y = r + s)} \).

Step 3: Calculate the Joint Probability

\( P(X = r, Y = s) = P(X = r) \cdot P(Y = s) \) because \( X \) and \( Y \) are independent. Since \( X \sim \text{Binomial}(n, p) \) and \( Y \sim \text{Binomial}(m, p) \), we know:\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]\[ P(Y = s) = \binom{m}{s} p^s (1-p)^{m-s} \]Thus, \( P(X = r, Y = s) = \binom{n}{r} \binom{m}{s} p^{r+s} (1-p)^{n+m-(r+s)} \).

Step 4: Calculate the Denominator for the Conditional Probability

The event \( X + Y = r + s \) can occur if \( X = r, Y = s \). Hence, the probability is derived as:\[ P(X + Y = r + s) = \sum_{k=0}^{r+s} P(X = k, Y = r+s-k) \]Summing over all possible pairs \( (k, r+s-k) \) would just yield:\[ \binom{n+m}{r+s} p^{r+s} (1-p)^{n+m-(r+s)} \]

Step 5: Simplify the Conditional Probability

Substitute the expressions from Step 3 (numerator) and Step 4 (denominator) into the conditional probability formula:\[ P(X = r \mid X + Y = r + s) = \frac{\binom{n}{r} \binom{m}{s} p^{r+s} (1-p)^{n+m-(r+s)}}{\binom{n+m}{r+s} p^{r+s} (1-p)^{n+m-(r+s)}} \]Upon simplification, the \( p^{r+s} (1-p)^{n+m-(r+s)} \) terms cancel out yielding:\[ P(X = r \mid X + Y = r + s) = \frac{\binom{n}{r} \binom{m}{s}}{\binom{n+m}{r+s}} \]

Step 6: Select the Correct Answer From the Options

The final simplified expression, \( \frac{\binom{n}{r} \binom{m}{s}}{\binom{n+m}{r+s}} \), does not match options (A), (B), or (C) exactly. Thus, the correct answer choice is (D) none of these.

Key Concepts

Conditional Probability

Conditional probability is a way to find the probability of an event happening, given that another event has already occurred. This concept is critical when dealing with problems involving random variables. Mathematically, the conditional probability of event A given event B is expressed as \( P(A \mid B) \). This can be calculated using the formula:

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

Here, \( P(A \cap B) \) is the probability that both events A and B happen. However, to find perfect clarity, always ensure \( P(B) eq 0 \) since it is the denominator.
In our exercise, we calculate the conditional probability of \( X = r \) given that \( X + Y = r + s \). This means we need to know the joint probability of both events occurring and the probability of the condition happening (i.e. \( X + Y = r + s \)). Understanding conditional probability helps us determine the likelihood of one random variable's value when another related condition is known to be true. This technique is utilized frequently in statistics and probability problems.

Independent Random Variables

Independent random variables are random variables that do not affect each other's outcomes. When two random variables, like \( X \) and \( Y \), are independent, the occurrence of a particular event involving one variable does not change or inform the probability of events involving the other.

• If \( X \) and \( Y \) are independent, then \( P(X = a, Y = b) = P(X = a) \cdot P(Y = b) \).

This property is very useful when solving probability problems because it allows separate probabilities to be multiplied directly without any adjustments.
In the context of our exercise, since \( X \) and \( Y \) are independent, we could express the joint probability \( P(X = r, Y = s) \) as the product of individual probabilities, \( P(X = r) \) and \( P(Y = s) \). This greatly simplifies calculations because it breaks down complex joint probabilities into simpler single-variable probabilities.

Law of Total Probability

The Law of Total Probability is an essential principle in probability theory that provides a way to calculate the total probability of an outcome which can be broken down into several mutually exclusive events. The law states:

• \( P(B) = P(B \cap A_1) + P(B \cap A_2) + \ldots + P(B \cap A_n) \),

where \( A_1, A_2, \ldots, A_n \) are various events and all their outcomes cover the sample space.
In our exercise, this law is indirectly used when identifying all ways \( X + Y = r + s \) can happen. We need to consider every possible value \( k \) that \( X \) could take, for which \( Y = r + s - k \), and then sum these probabilities, i.e., the probability of all combinations that sum up to \( r + s \). This approach is crucial for breaking down the complex scenarios covering combinations of two random variables into manageable parts, enabling us to find the denominator for the conditional probability calculation.