Q.Three coins are tossed simultaneously. Consider the event $E$ 'three heads or three tails', $F$ 'at least two heads' and $G$ 'at most two heads'. Of the pairs $(E,F)$, $(E,G)$ and $(F,G)$, which are independent? which are dependent?

Idea: Two events are independent iff $P(X\cap Y)=P(X)\,P(Y)$. Compute each probability from the 8 equally likely outcomes.

Sample space: $\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$.

• $E$ (three heads or three tails) $=\{HHH,TTT\}$, $P(E)=\tfrac{2}{8}=\tfrac14$.
• $F$ (at least two heads) $=\{HHH,HHT,HTH,THH\}$, $P(F)=\tfrac{4}{8}=\tfrac12$.
• $G$ (at most two heads) $=$ all except $HHH$, $P(G)=\tfrac{7}{8}$.

Checks:

• $(E,F)$: $E\cap F=\{HHH\}$, $P=\tfrac18$; $P(E)P(F)=\tfrac14\cdot\tfrac12=\tfrac18$. Equal $\Rightarrow$ independent. …

The test for independence

Events $X$ and $Y$ are independent exactly when

If the two sides differ, they are dependent. So we compute each probability from the sample space of tossing three fair coins:

eight equally likely outcomes, each of probability $\tfrac18$.

Probabilities of the three events

$E$ — three heads or three tails: $E=\{HHH,TTT\}$, so $P(E)=\dfrac{2}{8}=\dfrac14$.

$F$ — at least two heads (two or three heads): $F=\{HHH,HHT,HTH,THH\}$, so $P(F)=\dfrac{4}{8}=\dfrac12$.

$G$ — at most two heads (zero, one or two heads): this is everything except $HHH$, so $P(G)=\dfrac{7}{8}$.

Pair $(E,F)$

$E\cap F$ needs "three heads or three tails" AND "at least two heads." Only $HHH$ qualifies ($TTT$ has no heads), so $E\cap F=\{HHH\}$ and $P(E\cap F)=\dfrac18$.

Compare: $P(E)P(F)=\dfrac14\cdot\dfrac12=\dfrac18$. The two sides are equal, so $(E,F)$ is independent.

Pair $(E,G)$

$E\cap G$ needs "three heads or three tails" AND "at most two heads." Since $HHH$ is excluded by $G$, only $TTT$ remains, so $E\cap G=\{TTT\}$ and $P(E\cap G)=\dfrac18$. …

Method: Testing several event-pairs for independence

Use this when a question defines three events and asks which pairs are independent — apply the product test once per pair.

Steps

Step 1: Compute every marginal probability from a common sample space.

List the full equally-likely space (eight outcomes for three coins) and read off $P(E)$, $P(F)$, $P(G)$.

Step 2: For each pair, find the joint probability by intersecting the sets.

Determine $E\cap F$, $E\cap G$, $F\cap G$ as explicit outcome sets, then their probabilities. …

Common Mistakes

Mistake 1: Judging independence from overlap size rather than the product rule.

Why it's wrong: $(E,F)$ share only one outcome yet are independent, while $(F,G)$ share three yet are dependent — overlap alone tells you nothing. Correct approach: test each pair with $P(X\cap Y)=P(X)P(Y)$.

Mistake 2: Assuming that if one pair is independent the others are too. …