Question:
Let \(A,B,C\) be three events such that \(P(C)P(C^c)>0\). Consider the following statements.
Let \(A,B,C\) be three events such that \(P(C)P(C^c)>0\). Consider the following statements.
Show Hint
Use the law of total probability whenever conditional probabilities with respect to \(C\) and \(C^c\) are given together.
Updated On: Jun 4, 2026
- \(S1\) is correct and \(S2\) is NOT correct
- \(S1\) is NOT correct and \(S2\) is correct
- Both \(S1\) and \(S2\) are correct
- Neither \(S1\) nor \(S2\) is correct
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Use the law of total probability for statement \(S1\).
Since \(P(C)P(C^c)>0\), both \(P(C)>0\) and \(P(C^c)>0\).
Now, by the law of total probability,
\[ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \] and
\[ P(B)=P(B|C)P(C)+P(B|C^c)P(C^c) \]
Step 2: Compare \(P(A)\) and \(P(B)\).
Given that
\[ P(A|C)>P(B|C) \] and
\[ P(A|C^c)>P(B|C^c) \] Multiplying the first inequality by \(P(C)>0\), we get
\[ P(A|C)P(C)>P(B|C)P(C) \] Multiplying the second inequality by \(P(C^c)>0\), we get
\[ P(A|C^c)P(C^c)>P(B|C^c)P(C^c) \] Adding both inequalities,
\[ P(A|C)P(C)+P(A|C^c)P(C^c) > P(B|C)P(C)+P(B|C^c)P(C^c) \] Therefore,
\[ P(A)>P(B) \] So, statement \(S1\) is correct.
Step 3: Analyze statement \(S2\).
Given that
\[ P(A|C)=P(A|C^c) \] Let this common value be \(k\).
So,
\[ P(A|C)=k \] and
\[ P(A|C^c)=k \]
Step 4: Find \(P(A)\).
Again using the law of total probability,
\[ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \] Substitute the common value \(k\):
\[ P(A)=kP(C)+kP(C^c) \] \[ P(A)=k\{P(C)+P(C^c)\} \] Since
\[ P(C)+P(C^c)=1 \] we get
\[ P(A)=k \] Thus,
\[ P(A|C)=P(A) \]
Step 5: Conclude independence.
Since
\[ P(A|C)=P(A) \] and \(P(C)>0\), it follows that
\[ P(A\cap C)=P(A)P(C) \] Hence, \(A\) and \(C\) are independent.
Therefore, statement \(S2\) is correct.
Step 6: Final conclusion.
Both \(S1\) and \(S2\) are correct.
Hence,
\[ \boxed{(C)} \]
Since \(P(C)P(C^c)>0\), both \(P(C)>0\) and \(P(C^c)>0\).
Now, by the law of total probability,
\[ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \] and
\[ P(B)=P(B|C)P(C)+P(B|C^c)P(C^c) \]
Step 2: Compare \(P(A)\) and \(P(B)\).
Given that
\[ P(A|C)>P(B|C) \] and
\[ P(A|C^c)>P(B|C^c) \] Multiplying the first inequality by \(P(C)>0\), we get
\[ P(A|C)P(C)>P(B|C)P(C) \] Multiplying the second inequality by \(P(C^c)>0\), we get
\[ P(A|C^c)P(C^c)>P(B|C^c)P(C^c) \] Adding both inequalities,
\[ P(A|C)P(C)+P(A|C^c)P(C^c) > P(B|C)P(C)+P(B|C^c)P(C^c) \] Therefore,
\[ P(A)>P(B) \] So, statement \(S1\) is correct.
Step 3: Analyze statement \(S2\).
Given that
\[ P(A|C)=P(A|C^c) \] Let this common value be \(k\).
So,
\[ P(A|C)=k \] and
\[ P(A|C^c)=k \]
Step 4: Find \(P(A)\).
Again using the law of total probability,
\[ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \] Substitute the common value \(k\):
\[ P(A)=kP(C)+kP(C^c) \] \[ P(A)=k\{P(C)+P(C^c)\} \] Since
\[ P(C)+P(C^c)=1 \] we get
\[ P(A)=k \] Thus,
\[ P(A|C)=P(A) \]
Step 5: Conclude independence.
Since
\[ P(A|C)=P(A) \] and \(P(C)>0\), it follows that
\[ P(A\cap C)=P(A)P(C) \] Hence, \(A\) and \(C\) are independent.
Therefore, statement \(S2\) is correct.
Step 6: Final conclusion.
Both \(S1\) and \(S2\) are correct.
Hence,
\[ \boxed{(C)} \]
Was this answer helpful?
0
0
Top IIT JAM MS Statistics Questions
- A bag has 5 blue balls and 15 red balls. Three balls are drawn at random from the bag simultaneously. Then the probability that none of the chosen balls is blue equals
- Let \( Y \) be a continuous random variable such that \( P(Y > 0) = 1 \) and \( \mathbb{E}(Y) = 1 \). For \( p \in (0, 1) \), let \( \xi_p \) denote the \( p \)th quantile of the probability distribution of the random variable \( Y \). Then which of the following statements is always correct?
- Let 𝑋 be a continuous random variable having the 𝑈(−2, 3) distribution. Then which of the following statements is correct?
- Let \( X \) be a random variable having the Poisson distribution with mean \( 1 \). Let \( g: \mathbb{N} \cup \{0\} \to \mathbb{R} \) be defined by
\[g(x) = \begin{cases} 1 & \text{if } x \in \{0, 2\} \\ 0 & \text{if } x \notin \{0, 2\} \end{cases}\]
Then \( E(g(X)) \) is equal to: - For \( n \in \mathbb{N} \), let \( Z_n \) be the smallest order statistic based on a random sample of size \( n \) from the \( U(0,1) \) distribution. Let \( nZ_n \xrightarrow{d} Z \) as \( n \to \infty \) for some random variable \( Z \). Then \( P(Z \leq \ln 3) \) is equal to:
View More Questions
Top IIT JAM MS Probability Questions
- Let \( E \) and \( F \) be two events with \( 0<P(E)<1 \), \( 0<P(F)<1 \) and \( P(E) + P(F) \geq 1 \). Which of the following statements is (are) TRUE?
- The cumulative distribution function of a random variable \( X \) is given by
\[ F(x) = \begin{cases} 0, & \text{if } x < 0 \\ \frac{4}{9}, & \text{if } 0 \leq x < 1 \\ \frac{8}{9}, & \text{if } 1 \leq x < 2 \\ 1, & \text{if } x \geq 2 \end{cases} \] Which of the following statements is (are) TRUE?
- Let \( X_1, X_2 \) be a random sample from a distribution with the probability mass function
\[ f(x|\theta) = \begin{cases} 1 - \theta, & \text{if } x = 0 \\ \theta, & \text{if } x = 1 \\ 0, & \text{otherwise}, \quad 0 < \theta < 1. \end{cases} \] Which of the following is (are) unbiased estimator(s) of \( \theta \)?
- Let \( X_1, X_2, X_3 \) be a random sample from a distribution with the probability density function
\[ f(x|\theta) = \begin{cases} \frac{1}{\theta} e^{-x/\theta}, & \text{if } x > 0, \, \theta > 0 \\ 0, & \text{otherwise} \end{cases} \] If \( \hat{X}(X_1, X_2, X_3) \) is an unbiased estimator of \( \theta \), which of the following CANNOT be attained as a value of the variance of \( \hat{X} \) at \( \theta = 1 \)?
- Let \( X_1, X_2, \dots, X_n \) (where \( n \geq 2 \)) be a random sample from a distribution with the probability density function
\[ f(x|\theta) = \begin{cases} \frac{x}{\theta^2} e^{-x/\theta}, & x > 0, \, \theta > 0 \\ 0, & \text{otherwise} \end{cases} \] Let \( \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \). Which of the following statistics is (are) sufficient but NOT complete?
View More Questions
Top IIT JAM MS Questions
- Let \( a_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \) and \( b_n = \sum_{k=1}^{n} \frac{1}{k^2} \) for all \( n \in \mathbb{N} \). Then
- Let \( f(x,y) = x^2 + 3y^2 - \frac{2}{3} xy \) at \( (x,y) \in \mathbb{R}^2 \). Then
- Let \( A = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix} \) be a real matrix, where \( ad = 1 \) and \( c \ne 0 \). If \( A^{-1} + (\text{adj} \, A)^{-1} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \), then \( (\alpha, \beta, \gamma, \delta) \) is equal to
- A bag has 5 blue balls and 15 red balls. Three balls are drawn at random from the bag simultaneously. Then the probability that none of the chosen balls is blue equals
- Let \( Y \) be a continuous random variable such that \( P(Y > 0) = 1 \) and \( \mathbb{E}(Y) = 1 \). For \( p \in (0, 1) \), let \( \xi_p \) denote the \( p \)th quantile of the probability distribution of the random variable \( Y \). Then which of the following statements is always correct?
View More Questions