Math Examples of Conditional Probability

We use following steps to find the Probability that the card drawn is greater than 7 and less than 11-
Step 1: First of all, we assume event -
Let e₁ be the event for getting card greater than 7 and less than 11 and e₂ is the event of getting red card. Step 2: Now we need to find the probability of e1 when e2 is occurred, so we need to find p (e₁ / e₂) -
A pack of card contains 26 red cards and 26 black cards and we need so collect the black cards,
So p (e₁) = 26,
Now we need to find the black cards greater than 7 and less than 11.
As we know that in 26 black cards there will be 2 pairs of cards between 7 to 11. Now if we calculate then there will be 4 + 4 = 8 cards.
So p (e₂ ∩e₁) = 8
Now we will put that in the given formula we will get
P (e₁ / e₂) = 8 / 26,
P (e₁ / e₂) = 4 / 13,
This is the required probability for the given event. So, probability that the card drawn is greater than 7 and less than 11 is p (e₁ / e₂) = 4 / 13.

We use following steps for finding the probability of getting the Wednesday -
Step 1: First of all, we find out Conditional Probability means here given that the given day is Wednesday and given probability of getting Wednesday on that day is 0.7.
Therefore, probability of getting Wednesday on that day is 0.7.
Or P (Wednesday) = 0.7
Step 2: Now we calculate probability of getting the day is Wednesday and that a student is absent. Here given that probability of getting the day is Wednesday and that a student is absent is 0.07. So, probability of getting the day is Wednesday and that a student is absent is 0.07 or
P (Wednesday and absent) = 0.07,
Step 3: Now we apply conditional probability theorem on given problem -
P (absent | Wednesday) = P (Wednesday and absent) / P (Wednesday),
= 0.07 / 0.7,
= 0.10.
So, the probability that a student is absent given that today is Wednesday is 0.10 and there are 10% chances for getting the probability that a student is absent given that day is Wednesday.

Here we can easily see from the question that the Probability of occurrence of an event ‘B’ is 2 / 3.
So P (B) = 2 / 3
And the Intersection probability of event ‘A’ and ‘B’ is
P (A∩B) = 2 / 7,
Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
Probability of ‘A’ when the Event ‘B’ is already occurred
P (A / B) = P (A∩B) / P (B),
= (2 / 7) / (2 / 3),
= (6 / 14) = 3 / 7.
So the probability of ‘A’ when the even ‘B’ is already occurred is 3 / 7.

Here we can easily see from the question that the Probability of occurrence of an event ‘B’ is 1/ 5.
So P (B) = 1 / 5,
And the Intersection probability of event ‘A’ and ‘B’ is
P (A∩B) = 1 / 8,
Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
Probability of ‘A’ when the Event ‘B’ has already occurred,
P (A / B) = P (A∩B) / P (B),
= (1/ 8) / (1/ 5),
= ( 1 / 40).
So the probability of ‘A’ when the even ‘B’ is already occurred is 1 / 40.

Here we can easily see from the question that the Probability of occurrence an event ‘Y’ is 1 / 4.
So P(Y) = 1/ 4,
And the Intersection probability of event ‘X’ and ‘Y’ is
P (X ∩ Y) = 1 / 3,
Now according to the property of probability if two events ‘X’ and ‘Y’ occur then,
Probability of ‘X’ when the Event ‘Y’ has already occurred,
P(X / Y) = P (X∩Y) / P(Y),
= (1/ 3) / (1/ 4),
= (1 / 12).

Here we can easily see from the question that the Probability of occurrence an event B is 1 / 3
So P (B) = ( 1 / 3)
And the Intersection probability of event ‘A’ and ‘B’ is
P (A∩B) = (1 / 5),
Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
Probability of ‘A’ when the Event ‘B’ has already occurred
P (A / B) = P (A∩B) / P (B),
= (1/ 5) / (1/ 3),
= (3 / 5).
So the probability of ‘A’ when the event ‘B’ has already occurred is 3 / 5.

The Probability of students, who will pass and fail,
P (pass and fail) = .083,
Probability of students who will pass is
P (pass) = .65,
Now the probability of student who will fail
P (passes / fails) = P (pass and fail) / P (pass)
= 0.083 / 0.65 = 0.127 = 12 .7%.

Here we can clearly see that the Probabilities of Students are absent and the day is Tuesday.
P( Tuesday and Absent) = .006,
The Probability of Tuesday is,
P(Tuesday) = 0.2,
So the probability,
P(Absent/ Tuesday) = P(Tuesday and absent) / P(Tuesday),
= .03 / .2 = .15 = 15%.

Here we can easily see that the Probability of occurrence of an event ‘X’ is
P(X) = 3 / 8,
And the probability of occurring a event ‘B’ is
P(Y) = 5 / 8,
Intersection probability of event ‘X’ and ‘Y’ is
P (X∩Y) = 3 / 4,
Now we know from the probability property P(X∪Y) = P(X) + P(Y) – P(X∩Y)
= 3 / 8 + 5 / 8 – 3 / 4 = 1/ 4.
After finding the probability P(X ∪ Y) we use the property of probability in which
P (X' / Y') = 1 – P (X∩Y) / 1 – P(Y),
= 1- 3/ 4 / 1 – 5/8= (1 / 4) / 3 / 8 = 8 / 12 = 2 / 3.
So the probability P(X' / Y') = 2 / 3.

Here we can easily see that the Probability of occurring a event ‘A’ is P (A) =7 / 8,
And the probability of occurrence of an event ‘B’ is
P (B) = 3 / 8,
Intersection probability of event ‘A’ and ‘B’ is
P (A∩B) = 3 / 5,
According to the probability property P (A ∪ B) = P (A) + P (B) – P (A∩B),
Now we move P (A ∩ B) in left hand side and move P (A ∪ B) right hand side then,
P (A∩B) = P(A) + P(B) - P(A∪B),
After getting that we put the values in it. So
P (A∩B) = 7 / 8 + 3/ 8 – 3/ 4 = 5 / 4 - 3 / 5 = 13 / 20 = 13/ 20,
So the probability P (A ∩B) = 13/ 20. After that we put the value of P (A ∩B) in the property of probability which is
P (B/A) = P (A ∩B) / P (A) = (13/ 20) / (7/ 8) = 26 / 35.