Quoted from a popular book,

For instance, if we flip a fair coin twice, knowing whether the first flip is Heads gives us no information about whether the second flip is Heads. These events are independent.

On the other hand, knowing whether the first flip is Heads certainly gives us information about whether both flips are Tails. (If the first flip is Heads, then definitely it’s not the case that both flips are Tails.) These two events are dependent.

Mathematically, we say that two events E and F are independent if the probability that they both happen is the product of the probabilities that each one happens:

P(E,F) = P(E)P(F)

In the example above, the probability of “first flip Heads” is 1/2, and the probability of “both flips Tails” is 1/4, but the probability of “first flip Heads and both flips Tails” is 0.

QUESTION: In the example we say

P(first flip Heads) = 1/2
P(both flips Tails) = 1/4
P(first flip Heads and both flips Tails) = 1/2 * 1/4 (WHY NOT THIS?)
P(first flip Heads and both flips Tails) = 0 (CORRECT ANSWER)

Although it makes logical sense to put zero, but doesn't explain why doesn't it fit the formula

  • 2
    $\begingroup$ The formula doesn't apply because of course these events aren't independent. $\endgroup$
    – Mason
    Dec 12, 2018 at 4:37

3 Answers 3


You have already given yourself the answer.

The events $TT$ and $H\cdot$ are dependent. Independence of two events $E,F$ would mean that

  • $P(E|F)= \frac{P(E \cap F)}{P(F)}= P(E)$.

But in your case you have $$P(TT | H\cdot) = 0 \neq P(TT)$$


You are misinterpreting the expression a little bit.

When we say both flip heads, we are referring to $$P(H_1,H_2)=P(H_1)P(H_2)$$ which we can directly apply the formula.

But when we take it one step further with "first flip head and both flip tails" we are finding

$$P(H_1\wedge(T_1\wedge T_2))$$

for now the 'terms' $H_1$ and $(T_1\wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to $$P((H_1\wedge T_1)\wedge T_2))$$ which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate $$P((H_1\wedge T_1)\wedge T_2))=P(H_1\wedge T_1)P( T_2)$$ and clearly $$P(H_1\wedge T_1)=0$$ I hope this explains :)

  • $\begingroup$ Thanks for the answer and it explained the situation flawlessly. I won't ever forget this. $\endgroup$ Dec 12, 2018 at 8:02

Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.

\begin{align*} P(E) &= 1/2 \\ P(F) &= 1/4 \\ P(E,F) &= 0 \\ P(E,F) &\neq P(E) \cdot P(F) \\ \end{align*}

It doesn't fit the formula for independence because the two events are not independent.


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