Q.In a large metropolitan area, the probabilities are $.87, .36, .30$ that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

We are given $P(A \cup B) = 0.85$, $P(B) = 0.50$, and $P(A \cap B) = 0.30$. We need $P(A \cap B')$, the probability that $A$ occurs but $B$ does not.

Concept and intuition:

The event $A \cap B'$ is the part of $A$ that lies outside $B$. A classic way to find it is to first find $P(A)$ using the inclusion–exclusion formula for the union, then subtract the overlap $P(A \cap B)$. This works because $A$ is partitioned into two disjoint parts: $A \cap B$ and $A \cap B'$.

1. Use inclusion–exclusion to find $P(A)$. The formula for the union of two events is:

Substitute the known values:

Simplify:

So:

2. Find $P(A \cap B')$ using the partition of $A$. Since $A$ is the union of the disjoint events $A \cap B$ and $A \cap B'$, we have:

Rearranging:

Step 1 — Write what's given

$P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.

Step 2 — Find $P(A) + P(B)$

By the addition theorem of probability, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, so

$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.6 + 0.2 = 0.8$ …

Step 1 — Set up the sample space.

A die has 6 faces, and rolling it fairly makes each face equally likely. The stem tells us how the numbers are distributed across those faces:

(Good — the counts add to 6, so the description is consistent.)

The crucial subtlety: the equally likely outcomes are the six FACES, not the three distinct numbers. The numbers are not equally likely, because different numbers occupy different numbers of faces. Falling for "there are 3 possible numbers, two of them are favourable, so $2/3$" is the trap that produces option (A).

Step 2 — Find the individual probabilities.

Using the classical definition $P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$:

(For completeness, $P(2) = \tfrac{3}{6}$, and $\tfrac{2}{6}+\tfrac{3}{6}+\tfrac{1}{6} = 1$ ✓ — the probabilities sum to 1, confirming our model.)

Step 3 — The concept: addition rule for mutually exclusive events.

A single roll cannot show a '1' and a '3' at the same time — the events are mutually exclusive, so $P(1 \cap 3) = 0$. The addition rule

therefore reduces to simply

Step 4 — Compute.

$P(1 \text{ or } 3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}.$ …

Step 1 — Set up. Let $L$ = 'loaded' and $R$ = 'red'. Given $P(L\cap R)=0.2$ and $P(L\cup R)=0.6$ (a basket is either loaded or red 60 times per 100).

Step 2 — Inclusion–exclusion.

$0.6=P(L)+P(R)-0.2\;\Rightarrow\;P(L)+P(R)=0.8$ …

1. Conditional probability definition.

2. Find $P(A\cap B)$ from the first relation, using $P(A)=\tfrac14$ and $P(B\mid A)=\tfrac23$:

3. Find $P(B)$ from the second relation, using $P(A\mid B)=\tfrac12$: …

Step 1 — Name the events.

Let

• $L$ = there is a lockdown, so $L'$ = there is no lockdown,
• $C$ = the pandemic is controlled within one month.

Step 2 — Extract the given data.

Step 3 — Why the total-probability theorem applies.

$L$ and $L'$ are mutually exclusive and exhaustive — exactly one of them must happen. They therefore partition the sample space, and the event $C$ can only occur through one of these two routes. The theorem of total probability then says

(Intuitively: 'weight each pathway's success rate by how likely that pathway is, then add.')

Step 4 — Substitute.

$P(C) = (0.7)(0.8) + (0.3)(0.3)$ …

Concept: Addition rule for independent events.

P(A fails) = 0.2 , P(B fails) = 0.3 ; the two events are independent.

P(A and B both fail) = 0.2 * 0.3 = 0.06 …

The key idea here is the Addition Rule for mutually exclusive events: if events cannot happen together, the probability that any of them occurs is just the sum of their individual probabilities. And "exhaustive" means that together they cover all possible outcomes — so that sum must equal 1.

Let’s unpack the relation $P(A) = 2P(B) = 3P(C)$. This is a chain of equalities. It tells us that $P(A)$ is twice $P(B)$, and also three times $P(C)$. So we can write everything in terms of a single variable — the most natural choice is $P(B)$.

1. From $P(A) = 2P(B)$, we have $P(A) = 2P(B)$.
2. From $2P(B) = 3P(C)$, we get $P(C) = \frac{2}{3}P(B)$.

Now, because A, B, C are mutually exclusive and exhaustive:

Substitute the expressions:

Combine the terms. Write $2P(B) = \frac{6}{3}P(B)$ and $P(B) = \frac{3}{3}P(B)$, so:

$\frac{11}{3}P(B) = 1$ …

Step 1 — Write the binomial law.

For $X\sim B(n=6,p)$, with $q=1-p$:

Step 2 — Use the given data as a ratio.

(The printed values 12 and 5 cannot be probabilities — a probability never exceeds 1 — so the only meaningful reading is that they are in the ratio $12:5$.)

Step 3 — Solve.

With $p+q=1$:

Step 4 — Match the options. …