Q.Determine $P(E|F)$. A coin is tossed three times, where

Concept: Conditional Probability — $P(E|F) = \frac{P(E \cap F)}{P(F)}$, provided $P(F) > 0$.

(i) Sample space has 8 equally likely outcomes.

$F = \{HHH, HHT\}$, so $P(F) = \frac{2}{8} = \frac{1}{4}$.

$E \cap F = \{HHH\}$, so $P(E \cap F) = \frac{1}{8}$. …

The Core Idea: Conditional Probability

When we write $P(E|F)$, we are asking: If we already know that $F$ has happened, what is the chance that $E$ also happens? The key is that the sample space shrinks — we only consider outcomes where $F$ is true. Then $P(E|F)$ is simply the fraction of those $F$-outcomes that also belong to $E$.

Mathematically:

where the second equality holds when all outcomes are equally likely (as they are with a fair coin).

Let’s work through each part.

(i) $E$: head on third toss, $F$: heads on first two tosses

Step 1: List the sample space.

Tossing a coin three times gives $2^3 = 8$ equally likely outcomes:

Step 2: Identify $F$.

$F$ = heads on first two tosses. That means the first two positions are both H. The third can be anything. So:

Only 2 outcomes.

Step 3: Identify $E \cap F$.

$E$ = head on third toss. Among $F$, which outcomes have a head on the third toss? Only $HHH$. So:

That’s 1 outcome.

Step 4: Compute $P(E|F)$.

(ii) $E$: at least two heads, $F$: at most two heads

Step 1: List outcomes for $F$.

“At most two heads” means 0, 1, or 2 heads. That’s every outcome except the one with 3 heads ($HHH$). So:

That’s 7 outcomes.

Step 2: List outcomes for $E$.

“At least two heads” means 2 or 3 heads:

That’s 4 outcomes.

Step 3: Find $E \cap F$.

We need outcomes that are in both $E$ and $F$. Since $F$ excludes $HHH$, the intersection is:

That’s 3 outcomes.

Step 4: Compute $P(E|F)$.

$P(E|F) = \frac{3}{7}$ …

Method: Conditional probability by counting a coin-toss sample space

Use this whenever a fair coin is tossed a fixed number of times and you must find $P(E\mid F)$ for events described in words such as "head on the third toss", "at least two heads", or "at most two tails".

Steps

Step 1: Write the equally-likely sample space.

Tossing a coin $n$ times gives $2^n$ equally likely ordered outcomes (for $n=3$ that is the eight strings $HHH, HHT, \dots, TTT$). Because every outcome has the same probability, you may work with counts instead of probabilities:

Step 2: Translate each verbal event into an outcome set.

Convert the words carefully — "at least two heads" = $2$ or $3$ heads, "at most two heads" = every outcome except all-heads, "at least one tail" = every outcome except all-heads. Listing (or using the complement) avoids miscounting. …

Common Mistakes

Mistake 1: In part (ii), dividing by 8 (the full sample space) instead of 7.

Why it's wrong: once $F$ (at most two heads) is given, $HHH$ becomes impossible, so the world shrinks to $7$ outcomes, not $8$. Correct approach: use $P(E|F)=\frac{n(E\cap F)}{n(F)}=\frac{3}{7}$, never $\frac{n(E\cap F)}{8}$.

Mistake 2: Counting $HHH$ inside $E\cap F$ in part (ii).

Why it's wrong: $HHH$ has three heads, so it fails "at most two heads" and is not in $F$; including it gives a wrong $\frac{4}{7}$. Correct approach: $E\cap F=\{HHT,HTH,THH\}$ only, so the count is $3$. …