Q.Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.

Step 1 — Add the probabilities (mutually exclusive events)

Since $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$, so:

Step 2 — Take the complement …

Concept & Intuition

Mutually exclusive events cannot happen at the same time, so $P(A \cap B) = 0$. The key rules are:

• Every probability lies in $[0,1]$, so $0 \le P(A) \le 1$ and $0 \le P(B) \le 1$.
• For mutually exclusive events, $P(A \cup B) = P(A) + P(B) \le 1$ (since the union cannot exceed certainty).

We treat $x$ as a real number and find all $x$ satisfying these three inequalities simultaneously.

Step-by-step solution

1. Apply $0 \le P(A) \le 1$

So $0 \le x + \frac{1}{3} \le 1$.

• Lower bound: $x + \frac{1}{3} \ge 0 \implies x \ge -\frac{1}{3}$.
• Upper bound: $x + \frac{1}{3} \le 1 \implies x \le \frac{2}{3}$. Thus from $P(A)$ alone: $-\frac{1}{3} \le x \le \frac{2}{3}$.

2. Apply $0 \le P(B) \le 1$

So $0 \le \frac{1-x}{4} \le 1$. Multiply by 4: $0 \le 1-x \le 4$.

• Lower bound: $1-x \ge 0 \implies x \le 1$.
• Upper bound: $1-x \le 4 \implies -x \le 3 \implies x \ge -3$. Thus from $P(B)$ alone: $-3 \le x \le 1$.

3. Combine the two probability-range constraints Intersect $[-\frac{1}{3}, \frac{2}{3}]$ from step 1 with $[-3, 1]$ from step 2:

4. Apply the mutual exclusivity condition: $P(A) + P(B) \le 1$

Compute:

Require $\frac{3x}{4} + \frac{7}{12} \le 1$.

Subtract $\frac{7}{12}$: $\frac{3x}{4} \le 1 - \frac{7}{12} = \frac{5}{12}$. …

Let $P(A)=p$. Then $P(B)=\tfrac32p$ and $P(C)=\tfrac12P(B)=\tfrac34p$.

Exhaustive and mutually exclusive:

$p+\tfrac32p+\tfrac34p=1\Rightarrow p\left(\tfrac{4+6+3}{4}\right)=1\Rightarrow p\cdot\tfrac{13}{4}=1\Rightarrow p=\tfrac{4}{13}.$ …

Concept: For mutually exclusive and exhaustive events, P(A) + P(B) + P(C) = 1.

Setup: Let P(A) = 2P(B) = 3P(C) = k.

Then P(A) = k, P(B) = k/2, P(C) = k/3.

Step 1 — Use exhaustiveness:

k + k/2 + k/3 = 1

k(6 + 3 + 2)/6 = 1 → 11k/6 = 1 → k = 6/11.

Step 2 — Get P(B): …

Concept: for mutually exclusive and exhaustive events, P(A) + P(B) + P(C) = 1, and P(A union C) = P(A) + P(C).

Let P(A) = x.

P(B) = (3/2)x.

P(C) = (1/2)P(B) = (1/2)(3/2)x = (3/4)x.

Exhaustive: x + (3/2)x + (3/4)x = 1

=> (4x + 6x + 3x)/4 = 1

=> 13x/4 = 1 => x = 4/13. …

Step 1 — Recall the definition

Events $E$ and $F$ are mutually exclusive if $E \cap F = \varnothing$: they share no common outcome and cannot occur simultaneously.

Step 2 — Test each pair

• (A) Age 13 and in college: college requires roughly 17+ years, so both cannot be true together → mutually exclusive.
• (B) In school and a playback singer: perfectly compatible.
• (C) Elder brother and father has more than one son: in fact one implies the other; compatible.
• (D) Studied B.A. English and can read English novels: compatible (even reinforcing).

Step 3 — Select

Only (A) describes events that cannot coexist. …

The key idea here is the meaning of "mutually exclusive." Two events are mutually exclusive if they cannot happen at the same time — there is no overlap between them. In probability terms, that means $P(A \cap B) = 0$.

When you want the probability that at least one of the events happens (A or B), you use the addition rule:

Since $P(A \cap B) = 0$ for mutually exclusive events, this simplifies beautifully to:

Now, "neither A nor B" means that none of the events happen. That is the complement of "at least one happens." So:

Let's work it through:

1. Find $P(A \cup B)$ Since A and B are mutually exclusive:

2. Find the probability of neither $P(\text{neither}) = 1 - 0.8 = 0.2$ …

Step 1 — The general addition theorem.

For any two events,

The subtraction is there because outcomes lying in both $A$ and $B$ would otherwise be counted twice.

Step 2 — Use the "mutually exclusive" condition.

Mutually exclusive (disjoint) means $A$ and $B$ cannot occur together:

There is no double counting, so the theorem collapses to

Step 3 — Substitute the given values.

$P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} = 0.8$ …