Q.In an online jackpot, there is one first prize of ₹3,00,000, two second prizes of ₹2,00,000 each and three third prizes of ₹50,000 each. A total of 1,00,000 jackpot tickets, each costing ₹100, were sold. Rohan bought one ticket. Based on the above information, answer the following questions:

Let's break down this problem by first understanding the core concepts involved: conditional probability and the complement rule.

Conditional probability, $P(X|Y)$, represents the probability of event $X$ occurring given that event $Y$ has already occurred. Its definition is fundamental:

The complement rule states that the probability of an event not happening is $1$ minus the probability of it happening. If $X'$ denotes the complement of event $X$ (i.e., $X$ does not occur), then:

We are asked to find $P(A'|B')$, which means "the probability that event $A$ does not occur, given that event $B$ does not occur."

Now, let's work through the problem step-by-step.

1. Apply the definition of conditional probability. Using the formula $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, we replace $X$ with $A'$ and $Y$ with $B'$.

The problem states $P(B) \ne 1$. This is important because it implies $P(B') = 1 - P(B) \ne 0$, ensuring that the denominator is not zero and the conditional probability is well-defined.

2. Simplify the numerator using De Morgan's Law.

The term $A' \cap B'$ represents the event where neither $A$ nor $B$ occurs. This is equivalent to the event that $A \cup B$ (either $A$ or $B$ or both occur) does not occur. This is a direct application of De Morgan's Law for sets:

Therefore, we can rewrite the numerator:

3. Apply the complement rule to the numerator. Now we have $P((A \cup B)')$. Using the complement rule $P(X') = 1 - P(X)$, where $X$ is the event $(A \cup B)$:

4. Substitute back into the conditional probability formula. Substitute the simplified numerator back into the expression from Step 1: …

The heart of this problem is understanding what conditional probability means when both events are complemented, and how set operations interact with complements.

Conditional probability $P(A \mid B)$ asks: "Given that $B$ has occurred, what is the probability of $A$?" The formula is always

When we want $P(E' \mid F')$, we're asking: "Given that $F$ did not occur, what is the probability that $E$ also did not occur?" So we need the intersection $E' \cap F'$ in the numerator.

The key insight is recognizing what $E' \cap F'$ represents. By De Morgan's law, the region where neither $E$ nor $F$ occurs is precisely the complement of their union:

This transforms our problem into something we can express in terms of $P(E \cup F)$.

Step-by-step derivation:

1. Write the definition of conditional probability for $P(E' \mid F')$:

2. Apply De Morgan's law to the numerator: The event "not $E$ and not $F$" is the same as "not ($E$ or $F$)":

3. Express the complement in terms of probability:

4. Substitute back into the conditional probability formula:

This matches option (D) exactly.

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By De Morgan's law, $P(\text{not }A\text{ and not }B) = P(A'\cap B') = P\big((A\cup B)'\big) = 1 - P(A\cup B)$.

Using the addition rule: …

Why the complement works

When a problem asks for "at least one" of something, you face the tedious task of adding up many cases: exactly one head, exactly two heads, exactly three heads, or all four heads. The complement sidesteps this entirely. There's only one way to fail the "at least one head" condition: get zero heads, meaning all four tosses land tails. Calculate that single probability, subtract from 1, and you're done.

Solution

1. Identify the complement event

   "At least one head" means one or more heads. The opposite is "no heads at all," which is the same as getting tails on every single toss.

2. Find the probability of all tails

   Each toss is independent, and for an unbiased coin $P(\text{Tail}) = \frac{1}{2}$. The probability of tails on all four tosses is: …