Q.Determine $P(E|F)$. Two coins are tossed once, where

Use $P(E|F)=\dfrac{P(E\cap F)}{P(F)}$ on the sample space $\{HH,HT,TH,TT\}$.

(i) $E$: a tail on one coin, $F$: one coin shows a head. Here $F=\{HT,TH\}$ and $E=\{HT,TH\}$, so $E\cap F=\{HT,TH\}$.

$P(E|F)=\frac{2/4}{2/4}=1.$ …

The tool

With equally likely outcomes, restrict to $F$ and count:

The sample space for two coins is $\{HH,HT,TH,TT\}$, all equally likely.

Part (i): E = tail on one coin, F = one coin shows a head

• $F$ (exactly one head) $=\{HT,TH\}$, so $|F|=2$.
• $E$ (exactly one tail) is the same set $\{HT,TH\}$, so $E\cap F=\{HT,TH\}$ and $|E\cap F|=2$.

For two coins, 'exactly one head' and 'exactly one tail' describe the identical outcomes, so once $F$ happens, $E$ is certain.

Part (ii): E = no tail, F = no head …

Method: Restricted-count conditional probability, and spotting the extreme values 0 and 1

Use this for a small equally-likely experiment (here two coins) where you must find $P(E\mid F)$ and the events may turn out to coincide or to be mutually exclusive.

Steps

Step 1: List the equally-likely sample space.

Two coins give the four outcomes $\{HH, HT, TH, TT\}$, each of probability $\tfrac14$. With equally-likely outcomes,

Step 2: Write $E$ and $F$ as explicit outcome sets.

Turn each phrase into the exact outcomes it allows — for example "exactly one head" is $\{HT, TH\}$, "no tail" is $\{HH\}$, "no head" is $\{TT\}$.

Step 3: Compare the two sets before dividing.

Inspect the relationship, because it often gives the answer instantly: …

Common Mistakes

Mistake 1: Reporting $P(E)$ instead of $P(E|F)$ in part (i).

Why it's wrong: $P(E)=\frac{2}{4}=\frac12$, but conditioning on $F=\{HT,TH\}$ shrinks the world to just those two outcomes, both of which lie in $E$, so $P(E|F)=1$. Correct approach: divide by $|F|$, not by the full $4$.

Mistake 2: Assuming part (ii) must have a non-zero answer. …