Q.Events A and B are such that $P(A) = \frac{1}{2}$, $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$. State whether A and B are independent ?

We are told that $A$ and $B$ are independent events. This is the key that unlocks the problem: for independent events, the probability of both occurring is simply the product of their individual probabilities. We are also given the probability that neither occurs — that is, the complement of their union. Let’s denote $p = P(A)$ and $q = P(B)$. Our goal is to find $p$ and $q$.

1. Translate the given information into equations

• Simultaneous occurrence: $P(A \cap B) = \frac{1}{8}$. Because $A$ and $B$ are independent,

• Neither occurs: $P(\text{neither } A \text{ nor } B) = P(A^c \cap B^c) = \frac{3}{8}$. For independent events, the complements are also independent, so

So we have the system:

2. Expand the second equation

Substitute $pq = \frac{1}{8}$:

Simplify:

Thus:

3. Solve for $p$ and $q$

We now have:

These are the sum and product of the roots of a quadratic equation. The numbers $p$ and $q$ satisfy:

Multiply through by 8 to clear denominators:

$8x^2 - 6x + 1 = 0.$ …

For independent $A,B$: $P(A\cup B)=P(A)+P(B)-P(A)P(B)=\tfrac14+P(B)-\tfrac14P(B)$.

Given $P(A\cup B)=2P(B)-P(A)=2P(B)-\tfrac14$. Equate:

$\tfrac14+P(B)-\tfrac14P(B)=2P(B)-\tfrac14.$ …

Step 1 — The concept: addition rule + independence

The general addition rule is

When $A$ and $B$ are independent, the occurrence of one does not change the chance of the other, so

Combining the two:

With $P(B) = x$ this becomes a single linear equation in $x$:

Step 2 — A necessary consistency check on the data

Because $A \subseteq A\cup B$, we must always have $P(A\cup B) \ge P(A)$. The stem prints $P(A) = 0.75$ and $P(A\cup B) = 0.65$, i.e. $P(A\cup B) < P(A)$ — impossible. Solving with those numbers confirms it:

a negative probability, which cannot be. So $P(A)$ must read $\mathbf{0.25}$ (the value consistent with $P(A\cup B)=0.65$) — and, as Step 3 shows, $0.25$ is precisely the value that reproduces one of the printed options exactly.

Step 3 — Solve

With $P(A) = 0.25$, $P(A\cup B) = 0.65$, $P(B) = x$: …

Step 1 — Verify Statement (I): if $E$ and $F$ are independent, so are $E'$ and $F'$.

Given independence: $P(E \cap F) = P(E)\,P(F)$.

Use De Morgan's law, $E' \cap F' = (E \cup F)'$:

Apply the addition theorem, then the independence hypothesis:

Now factor by grouping:

Since $P(E' \cap F') = P(E')P(F')$, the complements $E'$ and $F'$ are independent.

$\Rightarrow$ Statement (I) is TRUE.

(Intuition: independence means knowing whether $E$ happened tells you nothing about $F$. Knowing whether $E$ didn't happen is the same information, so it still tells you nothing about $F$ — or about $F$ not happening.)

Step 2 — Verify Statement (II): two mutually exclusive events with non-zero probabilities cannot be independent.

Suppose $E$ and $F$ are mutually exclusive with $P(E) > 0$ and $P(F) > 0$.

Mutually exclusive means they cannot occur together:

But independence would require …

Concept & Intuition

Odds are a different way of expressing probability. "Odds against Arjun" means the ratio of failure to success; "odds in favour of Bhavana" means the ratio of success to failure. To find the probability that neither solves the problem, we first convert each person's odds into a probability of failure, then multiply them — assuming their attempts are independent.

1. Convert Arjun's odds to probability

   Odds against Arjun = $5:2$ means for every 5 failures, there are 2 successes.

   Total outcomes = $5 + 2 = 7$.

   Probability Arjun fails = $\frac{5}{7}$.

   Probability Arjun succeeds = $\frac{2}{7}$.

2. Convert Bhavana's odds to probability

   Odds in favour of Bhavana = $3:4$ means for every 3 successes, there are 4 failures.

   Total outcomes = $3 + 4 = 7$.

   Probability Bhavana fails = $\frac{4}{7}$.

   Probability Bhavana succeeds = $\frac{3}{7}$.

3. Probability neither solves

   The problem is not solved by either only if both fail. Assuming independence:

   $$$$

We are told the outcomes of the two matches are independent. For each match, the probabilities are:

• Jaipur wins: $\frac{1}{2}$
• Draw: $\frac{3}{10}$
• Jaipur loses: $\frac{1}{5}$

Points per match: win = 5, draw = 3, loss = 0.

We need the probability that after two matches, Jaipur’s total points are strictly greater than Delhi’s.

Concept and intuition:

Since each match gives points to both teams simultaneously, we can think of the difference in points per match.

• If Jaipur wins: Jaipur +5, Delhi +0 → difference = +5 for Jaipur.
• If draw: both get +3 → difference = 0.
• If Jaipur loses: Jaipur +0, Delhi +5 → difference = –5 for Jaipur.

After two matches, the total difference is the sum of the two per-match differences. Jaipur has more points if and only if this total difference is positive.

Because the matches are independent, we can list all 3×3 = 9 possible pairs of outcomes, compute the total difference, and sum probabilities where it is > 0.

Step-by-step solution:

1. List all possible outcomes for the two matches.

   Let the outcomes be (result of match 1, result of match 2). Each result can be W (Jaipur win), D (draw), or L (Jaipur loss).

   The probability of each pair is the product of the individual probabilities (independence).

2. Compute the point difference for each match:

   • W: difference = +5
   • D: difference = 0
   • L: difference = –5

   After two matches, total difference = sum of the two differences.

3. Enumerate all 9 cases and find where total difference > 0:

   Match 1	Match 2	Diff 1	Diff 2	Total diff	Jaipur > Delhi?
   W	W	+5	+5	+10	Yes
   W	D	+5	0	+5	Yes
   W	L	+5	–5	0	No
   D	W	0	+5	+5	Yes
   D	D	0	0	0	No
   D	L	0	–5	–5	No
   L	W	–5	+5	0	No

Concept and intuition

When two events are independent — meaning the outcome of the first doesn’t affect the second — the probability that both happen is simply the product of their individual probabilities. Here, each toss of a fair die is independent, so we can treat the two conditions separately and multiply.

1. First toss condition We want a 4, 5, or 6. That’s 3 favorable outcomes out of 6 equally likely faces.

2. Second toss condition We want a 1, 2, 3, or 4. That’s 4 favorable outcomes out of 6.

3. Combine using independence Since the two tosses are independent, $P(\text{both conditions}) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}.$ …