Given three indentical bags each containing $10$ balls, whose colours are as follows:
| | Red | Blue | Green |
|--------|-----|------|-------|
| Bag I | $3$ | $2$ | $5$ |
| Bag II | $4$ | $3$ | $3$ |
| Bag III| $5$ | $1$ | $4$ |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $(\frac{1}{p} + \frac{1}{q})$ is:
(A)
$6$
(B)
$9$
(C)
$7$
(D)
$8$
EASY
Correct Answer:C
Explanation
Probability that a Red ball comes from Bag I ($p$):
$p(B_1/R) = \frac{p(B_1) \cdot p(R/B_1)}{p(R)}$
Substituting the values,
$p(B_1/R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$
Simplify to find $p$:
$=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4}$
So, $p = \frac{1}{4}$.
Probability that a Green ball comes from Bag III ($q$):
$p(B_3/G) = \frac{p(B_3) \cdot p(G/B_3)}{p(G)}$
Substitute the values,
$p(B_3/G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}}$
Simplify to find $q$:
$=\frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3}$
So, $q = \frac{1}{3}$.
Calculation of $(\frac{1}{p} + \frac{1}{q})$:
$\frac{1}{p} = 4$, $\frac{1}{q} = 3$
Therefore,
$(\frac{1}{p} + \frac{1}{q}) = 4 + 3 = 7$
AI Tutor Explanation
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COMPETENCY
APPLY
MCQ_SINGLE
4 Mark(s)
2025
JEE Main 2025 (Online) 2nd April Evening Shift
Let $A = {1, 2, 3, ..., 100}$ and $R$ be a relation on $A$ such that $R = {(a, b) : a = 2b + 1}$. Let $(a_1, a_2), (a_2, a_3), (a_3, a_4), ..., (a_k, a_{k+1})$ be a sequence of $k$ elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k , for which such a sequence exists, is equal to :
(A)
6
(B)
8
(C)
7
(D)
5
MEDIUM
Correct Answer:A
Explanation
The relation $R$ is defined on the set $A = {1, 2, 3, ..., 100}$ such that $R = {(a, b) : a = 2b + 1}$. We need to find the largest integer $k$ for which there exists a sequence of $k$ ordered pairs from $R$ where the second element of each pair is the first element of the next pair.
The sequence in terms of $k$ is: $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$. Here, each $a_i$ satisfies the equation $a_i = 2a_{i+1} + 1$. Consequently, $a_1 = 2a_2 + 1$, making $a_1$ an odd number.
Let's examine the pattern:
$a_2 = 2a_3 + 1$, implying $a_1 = 2(2a_3 + 1) + 1 = 4a_3 + 3$.
$a_3 = 2a_4 + 1$, leading to $a_1 = 4(2a_4 + 1) + 3 = 8a_4 + 7$.
Continuing this pattern, we find:
$a_k = 2a_{k+1} + 1 \implies a_1 = 2^k \cdot a_{k+1} + (2^k - 1)$ where $a_{k+1}$ needs to be in set $A$. This implies:
$a_{k+1} = \frac{a_1 + 1 - 2^k}{2^k}$
Thus, $2^k \mid (a_1 + 1)$. The task is to find the highest $k$ where $2^k$ divides any $e_i$ in ${2, ..., 101}$.
The largest power of 2 that divides an element within this range determines $k$.
After computation, we find that $k$ can be a maximum of 6 because $2^6 = 64$ divides $95+1 = 96$, but $2^7 = 128$ does not divide any $e_i$ for $e_i \in A$. Therefore, the maximum $k$ is 6.
The sequence corresponding to this maximum $k$ is: $(95, 47), (47, 23), (23, 11), (11, 5), (5, 2)$