Question:
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
Updated On: Apr 12, 2026
- $\frac{7}{10}$
- $\frac{10}{17}$
- $\frac{12}{19}$
- $\frac{7}{19}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
This is a conditional probability problem that can be solved using Bayes' Theorem. We have two possible sources for a letter, and we are given some evidence (observing 'AN' on the envelope). We need to find the probability of one source given this evidence.
Step 2: Key Formula or Approach:
Let's define the events:
$E_1$: The letter is from KANPUR.
$E_2$: The letter is from ANANTPUR.
$A$: The event that two consecutive letters 'AN' are visible on the envelope.
We want to find $P(E_2 | A)$.
Bayes' Theorem states:
\[ P(E_2 | A) = \frac{P(A | E_2) P(E_2)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \] Step 3: Detailed Explanation:
First, we determine the prior probabilities. Since the letter could be from either place, it's reasonable to assume they are equally likely.
\[ P(E_1) = P(E_2) = \frac{1}{2} \] Next, we calculate the conditional probabilities of observing 'AN' given the source.
For KANPUR ($E_1$):
The word is KANPUR. The possible pairs of consecutive letters are:
KA, AN, NP, PU, UR.
There are a total of 5 pairs of consecutive letters.
The pair 'AN' appears 1 time.
So, the probability of observing 'AN' given the letter is from KANPUR is:
\[ P(A | E_1) = \frac{\text{Number of 'AN' pairs}}{\text{Total number of pairs}} = \frac{1}{5} \] For ANANTPUR ($E_2$):
The word is ANANTPUR. The possible pairs of consecutive letters are:
AN, NA, AN, NT, TP, PU, UR.
There are a total of 7 pairs of consecutive letters.
The pair 'AN' appears 2 times.
So, the probability of observing 'AN' given the letter is from ANANTPUR is:
\[ P(A | E_2) = \frac{\text{Number of 'AN' pairs}}{\text{Total number of pairs}} = \frac{2}{7} \] Now, we apply Bayes' Theorem:
\[ P(E_2 | A) = \frac{P(A | E_2) P(E_2)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \] \[ P(E_2 | A) = \frac{\left(\frac{2}{7}\right) \left(\frac{1}{2}\right)}{\left(\frac{1}{5}\right) \left(\frac{1}{2}\right) + \left(\frac{2}{7}\right) \left(\frac{1}{2}\right)} \] We can cancel the common factor of $1/2$ from the numerator and denominator:
\[ P(E_2 | A) = \frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}} \] Now, simplify the denominator:
\[ \frac{1}{5} + \frac{2}{7} = \frac{7 \times 1 + 5 \times 2}{5 \times 7} = \frac{7 + 10}{35} = \frac{17}{35} \] So, the probability is:
\[ P(E_2 | A) = \frac{\frac{2}{7}}{\frac{17}{35}} = \frac{2}{7} \times \frac{35}{17} = 2 \times \frac{5}{17} = \frac{10}{17} \] Step 4: Final Answer:
The probability that the letter came from ANANTPUR is $\frac{10}{17}$.
This is a conditional probability problem that can be solved using Bayes' Theorem. We have two possible sources for a letter, and we are given some evidence (observing 'AN' on the envelope). We need to find the probability of one source given this evidence.
Step 2: Key Formula or Approach:
Let's define the events:
$E_1$: The letter is from KANPUR.
$E_2$: The letter is from ANANTPUR.
$A$: The event that two consecutive letters 'AN' are visible on the envelope.
We want to find $P(E_2 | A)$.
Bayes' Theorem states:
\[ P(E_2 | A) = \frac{P(A | E_2) P(E_2)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \] Step 3: Detailed Explanation:
First, we determine the prior probabilities. Since the letter could be from either place, it's reasonable to assume they are equally likely.
\[ P(E_1) = P(E_2) = \frac{1}{2} \] Next, we calculate the conditional probabilities of observing 'AN' given the source.
For KANPUR ($E_1$):
The word is KANPUR. The possible pairs of consecutive letters are:
KA, AN, NP, PU, UR.
There are a total of 5 pairs of consecutive letters.
The pair 'AN' appears 1 time.
So, the probability of observing 'AN' given the letter is from KANPUR is:
\[ P(A | E_1) = \frac{\text{Number of 'AN' pairs}}{\text{Total number of pairs}} = \frac{1}{5} \] For ANANTPUR ($E_2$):
The word is ANANTPUR. The possible pairs of consecutive letters are:
AN, NA, AN, NT, TP, PU, UR.
There are a total of 7 pairs of consecutive letters.
The pair 'AN' appears 2 times.
So, the probability of observing 'AN' given the letter is from ANANTPUR is:
\[ P(A | E_2) = \frac{\text{Number of 'AN' pairs}}{\text{Total number of pairs}} = \frac{2}{7} \] Now, we apply Bayes' Theorem:
\[ P(E_2 | A) = \frac{P(A | E_2) P(E_2)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \] \[ P(E_2 | A) = \frac{\left(\frac{2}{7}\right) \left(\frac{1}{2}\right)}{\left(\frac{1}{5}\right) \left(\frac{1}{2}\right) + \left(\frac{2}{7}\right) \left(\frac{1}{2}\right)} \] We can cancel the common factor of $1/2$ from the numerator and denominator:
\[ P(E_2 | A) = \frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}} \] Now, simplify the denominator:
\[ \frac{1}{5} + \frac{2}{7} = \frac{7 \times 1 + 5 \times 2}{5 \times 7} = \frac{7 + 10}{35} = \frac{17}{35} \] So, the probability is:
\[ P(E_2 | A) = \frac{\frac{2}{7}}{\frac{17}{35}} = \frac{2}{7} \times \frac{35}{17} = 2 \times \frac{5}{17} = \frac{10}{17} \] Step 4: Final Answer:
The probability that the letter came from ANANTPUR is $\frac{10}{17}$.
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Probability Questions
- Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability that the ball drawn is white is \(\frac{29}{45}\), then n is equal to:
If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)
- Let \(X\) have a binomial distribution \(B(6,p)\). If the sum of the mean and the variance of \(X\) is \(\dfrac{21}{8}\), then \[ \frac{P(2\le X<4)}{P(4<X<6)} \] is equal to:
- Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3, \ldots, 50$. The probability that their product $ab$ is divisible by $3$ is
The probability distribution of a random variable \(X\) is given below:
If \(E(X)=\dfrac{263}{15}\), then \(P(X<20)\) is equal to:
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions