Conditional probability

Lesson 6 Chapter 2

One of the essential concepts in probability theory is the conditional probability. This is because a lot of events depends on other precedent events or available partial information. Recognizing and calculating this dependency can lead to a more precise probability estimation. As below figure, how many layers should you wear, depends on the weather!!

Conditional probability definition

Conditional probability, using simple wording, refers to the likelihood of an event (chain of events) given the fact that another event (chain of events) happened.

Let's have an example. Assume we toss a dice and then flip a fair coin. What is the probability of getting number one (Event $A$ ) in tossing the dice and getting heads (Event $B$ ) after flipping the coin? Clearly $A \cap B = \varnothing$ . So, the probability of getting number one (in tossing the dice) and getting heads (in flipping the coin) is as below:

$P(A \cap B) = P(A)P(B)={\frac{1}{6} \times \frac{1}{2}} = \frac{1}{12}$

Let's take a look at a conditional situation. Assume we want to calculate the probability of having heads (in flipping the coin) if we get one (in tossing the dice)? This is a conditional statement. Basically, $B$ is conditioned on happening $A$ :

$P(B|A)$

The mathematical formulation of the conditional probability of two events is as below:

Conditional probability calculation

If $P(A)>0$ , then the conditional probability of $P(B|A)$ is a as below:

$P(B|A)=\frac{P(A \cap B)}{P(B)}$

NOTE: In case $A \cap B = \varnothing$ (The two $A$ and $B$ events are independent), we have:

$P(B|A)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)P(B)}{P(B)}=P(A)$

Question

Assume we toss a dice twice. What is the probability that the summation of the two numbers is higher than 8 if we get number 4 in our first trial?

Solution: Let's denote events as follows: (A) summation of the two numbers is higher than eight, and (B) we get the number 4 in our first trial. Definitely, $A$ and $B$ are not independent, and we want to calculate $P(A|B)$ . The calculation of $P(B)$ is easy. As we may see, all numbers with equal probability in tossing a dice, then $P(B)=\frac{1}{6}$ . Now we need to calculate $P(A \cap B)$ . The event $E = A \cap B$ indicates that we already know that the first dice number was 4. What is the event space of $E$ ? In other words, what should we see as the number of the second dice that makes the summation higher than 8? The following pairs demonstrate the event space of $E$ as the first and second elements of the pairs are associated with the first and second tossing of the dice:

$E=\{(4,5),(4,6)\}$

The sample space is as below:

$\Omega=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),\ldots,(6,5)(6,6)\}$

So the event space has only two elements, and the sample space has 36 elements. So $P(A \cap B) = \frac{2}{36}=\frac{1}{18}$ . So the final answer would be as follows:

$P(B|A)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{18}}{\frac{1}{6}}=\frac{1}{3}$

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