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Independence 10/14 and 10/16

Definition:Two events E and F are said to be independent if

\begin{displaymath}P(E\vert F)=P(E)=P(E\vert F^c),\qquad P(F)>0\end{displaymath}

Example:
We draw two cards one at a time from a shuffled deck of 52 cards.

Are the two following events independent?

E
The first card is a heart.
F
The second card is a queen.

From the definition of conditional probability, we need to find P(F|E) by computing

\begin{displaymath}P(E\cap F)=P(1st\; is\;
\heartsuit \; and\; 2nd\; is\; Q )=P(\heartsuit,Q)=\frac{51}{51 \times
52}\end{displaymath}

and $P(E)=\frac{1}{4}$.

or we showed that equivalently, E and F are independent if and only if:

\begin{displaymath}P(E\cap F)=P(E)\times P(F)\end{displaymath}

Beware this multiplication rule is ONLY available if and only if the events ARE independent.

Example:
De Méré's problem is whether or not it is more likely to get at least one double six in 24 throws of a pair of dice or to get at least one six in 4 throws of a die?

Essential to this argument is the fact that each throw of a die is independent of the preceding one.

The easier probability to compute is the complementary one:


\begin{displaymath}P(no\; double\; sixes\; in\; twenty\; four\; throws)=
(\frac{35}{36})^{24}=.509\end{displaymath}

and the complementary event

\begin{displaymath}P(at\; least\; one\;double\;six)=.491\end{displaymath}


\begin{displaymath}P(at \; least\; one\; six\; in\; four\; throws)=1-P(no \; sixes\; in\; 4 \;throws)=
1-(\frac{5}{6})^{4}=1-.482=.518\end{displaymath}



 
next up previous index
Next: Independence of more than Up: Conditional Probability Previous: Monty Hall 10/14
Susan Holmes
1998-12-07