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NIMCET 2023 Mathematics PYQ — Given two events A and B such that odd in favour A are 2:1 and od… | Mathem Solvex

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NIMCET 2023 Mathematics PYQ — Given two events A and B such that odd in favour A are 2:1 and od… | Mathem Solvex | Mathem Solvex

Explanation

Step 1: Determine individual probabilities from the given odds

The relation between odds in favor ( $a : b$ ) and probability is given by $P (E) = \frac{a}{a + b}$ .

Given odds in favor of $A = 2 : 1$
$P (A) = \frac{2}{2 + 1} = \frac{2}{3}$
Given odds in favor of $A \cup B = 3 : 1$
$P (A \cup B) = \frac{3}{3 + 1} = \frac{3}{4}$

Step 2: Apply the addition theorem of probability

According to set theory and probability:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Substitute the known values into the equation:

$\frac{3}{4} = \frac{2}{3} + P (B) - P (A \cap B)$

Rearranging the formula to express $P (B)$ in terms of $P (A \cap B)$ :

$P (B) = \frac{3}{4} - \frac{2}{3} + P (A \cap B)$

$P (B) = \frac{9 - 8}{12} + P (A \cap B)$

$P (B) = \frac{1}{12} + P (A \cap B) — (Equation 1)$

Step 3: Establish constraints to find the minimum and maximum values

The intersection of two sets, $A \cap B$ , must satisfy standard set theory limits:

$0 \leq P (A \cap B) \leq min (P (A), P (B))$

Additionally, since $P (A \cup B) = \frac{3}{4}$ , the maximum possible value for any component probability set bounded within this union cannot exceed the union value itself.

For the Minimum Value of $P (B)$ :
The smallest possible value for the intersection probability is $P (A \cap B) = 0$ (when events $A$ and $B$ are mutually exclusive).
Substituting $P (A \cap B) = 0$ into Equation 1:
$P (B)_{m i n} = \frac{1}{12} + 0 = \frac{1}{12}$
For the Maximum Value of $P (B)$ :
The largest value occurs when $B$ expands as much as possible within the sample space constraints. Since the maximum upper bound of any probability is $1$ and it cannot exceed the boundaries defined by standard logic, we check the condition for the intersection maximum bound:
$P (A \cap B) \leq P (A)$
$P (A \cap B) \leq \frac{2}{3}$
Also, the maximum probability value for $P (A \cup B)$ cannot exceed the sum of individual subsets if they overlap completely, meaning $B$ can completely encapsulate the remainder of the union space up to $P (A \cup B) = \frac{3}{4}$ .
Thus, the absolute maximum probability for event $B$ cannot exceed the overall probability of the union of both events:
$P (B) \leq P (A \cup B)$
$P (B)_{m a x} = \frac{3}{4}$

Step 4: Combine the inequalities

Combining both the minimum and maximum boundary values calculated above:

$\frac{1}{12} \leq P (B) \leq \frac{3}{4}$

Correct Answer: A) $\frac{1}{12} \leq P (B) \leq \frac{3}{4}$

Explanation

Step 1: Determine individual probabilities from the given odds

The relation between odds in favor ( $a : b$ ) and probability is given by $P (E) = \frac{a}{a + b}$ .

Given odds in favor of $A = 2 : 1$
$P (A) = \frac{2}{2 + 1} = \frac{2}{3}$
Given odds in favor of $A \cup B = 3 : 1$
$P (A \cup B) = \frac{3}{3 + 1} = \frac{3}{4}$

Step 2: Apply the addition theorem of probability

According to set theory and probability:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Substitute the known values into the equation:

$\frac{3}{4} = \frac{2}{3} + P (B) - P (A \cap B)$

Rearranging the formula to express $P (B)$ in terms of $P (A \cap B)$ :

$P (B) = \frac{3}{4} - \frac{2}{3} + P (A \cap B)$

$P (B) = \frac{9 - 8}{12} + P (A \cap B)$

$P (B) = \frac{1}{12} + P (A \cap B) — (Equation 1)$

Step 3: Establish constraints to find the minimum and maximum values

The intersection of two sets, $A \cap B$ , must satisfy standard set theory limits:

$0 \leq P (A \cap B) \leq min (P (A), P (B))$

Additionally, since $P (A \cup B) = \frac{3}{4}$ , the maximum possible value for any component probability set bounded within this union cannot exceed the union value itself.

For the Minimum Value of $P (B)$ :
The smallest possible value for the intersection probability is $P (A \cap B) = 0$ (when events $A$ and $B$ are mutually exclusive).
Substituting $P (A \cap B) = 0$ into Equation 1:
$P (B)_{m i n} = \frac{1}{12} + 0 = \frac{1}{12}$
For the Maximum Value of $P (B)$ :
The largest value occurs when $B$ expands as much as possible within the sample space constraints. Since the maximum upper bound of any probability is $1$ and it cannot exceed the boundaries defined by standard logic, we check the condition for the intersection maximum bound:
$P (A \cap B) \leq P (A)$
$P (A \cap B) \leq \frac{2}{3}$
Also, the maximum probability value for $P (A \cup B)$ cannot exceed the sum of individual subsets if they overlap completely, meaning $B$ can completely encapsulate the remainder of the union space up to $P (A \cup B) = \frac{3}{4}$ .
Thus, the absolute maximum probability for event $B$ cannot exceed the overall probability of the union of both events:
$P (B) \leq P (A \cup B)$
$P (B)_{m a x} = \frac{3}{4}$

Step 4: Combine the inequalities

Combining both the minimum and maximum boundary values calculated above:

$\frac{1}{12} \leq P (B) \leq \frac{3}{4}$

NIMCET 2023 — Mathematics PYQ

Choose the correct answer:

Explanation

Correct Answer: A) $\frac{1}{12} \leq P (B) \leq \frac{3}{4}$

Explanation

Correct Answer: A) $\frac{1}{12} \leq P (B) \leq \frac{3}{4}$