Welcome to the interesting lesson of Probability in quantitative aptitude or numeric ability section.
We use probability in routine life daily, Probability is simply a chance of happening of some event. If one of you friend always come late, then the probability of him coming on time today is very less. This is a generic simple understanding, but in mathematical terms, we define the probability of occurrence of some event by a number between 0 and 1. Here 0 means the event will not occur and 1 means the event will definitely occur.
Before moving deeper, let me introduce you a term Sample Space(S). When we perform some experiment, then set of all the possible outcomes(results) is called Sample Space(S).
so, for coin Sample space S={H,T}
For two coins tossed simultaneously S={HH, HT, TH, TH}
For rolling a dice, S={1,2,3,4,5,6}
Simply sample space is a list of all the possible results, and Event is any one or more than one result from sample space. (Event = H, HH, HT, 1,3,...)
1. Probability of occurrence of single event
Let's take a simple example to understand the concept. Suppose you are tossing a coin, then there will be either head or tail. The probability of coming head or tail is 0.5.
Now probability = Number of ways an event can occur / Total Events or Sample Space
P(E) = n(E) / n(S)
P(Head)= 1/2 = 0.5
n(E) = Number of ways head can occur means 1.
n(S)= Total Events means 2, head or tail
Another thing to note P(head)0.5 + P(Tail) 0.5= 1
If we add the probability of all the events the answer should be 1.
Let's take another example of dice.
While rolling a dice, we have six outcomes S={1,2,3,4,5,6}. What is a chance of coming 2 on the top face while rolling a dice. Here a dice is unbiased, and each face is likely to have the same probability of occurrence, so the probability of any particular number is 1/6.
Here n(S)=6, and n(E)=1.
2. Probability of occurrence of two or more event in single experiment
So far we were trying to calculate the probability of a single outcome/event. What if we want to find the probability of two, three or more outcomes together?
In the above dice example, suppose we want to find the probability of getting a multiple of 3 on the top face. Multiple of 3 in the dice are 3 and 6. we can solve this in two ways,
One way is to find the n(E), here we have two results 3 & 6 so n(E) will be 2. while n(S) is 6. putting them in the probability equation,
P(E) = n(E)/n(S) = 2/6 =1/3
Another way to solve is, to add the individual probability of both the events 3 & 6.
Probability of Multiple of 3 = P(3) + P(6) = 1/6 + 1/6 = 2/6.
3. Probability of occurrence of events in two independent experiment
We have already learned how to calculate the probability of occurrence of events in the single experiment, withing next few minutes you will be able to master this topic.
Now suppose we take one unbiased coin and one dice, and while throwing them simultaneously we want to find the probability of coming head on the coin and 3 on the dice. We already know that probability of coming head is 1/2, and the probability of coming 3 on top face of the dice is 1/6.
Now that coin and dice are a completely independent event, means there is no relation between coming head and 3, to find the combined probability we will multiply the independent probabilities.
Probability of coming head and 3 = P(head) * P(3) = 1/2 * 1/6 = 1/12
Wow! So now you can not only add but also multiply probabilities. But how do you know when to add or multiply probabilities?
Well, in the first example we had to find the probability of coming 3 or 6 on the top face of dice. In this case, either event alone would have occurred. In such cases, we simply add the individual probabilities.
On the other hand, in the second example, both the events needed to occur (head and 3 simultaneously). In such cases, we need to multiply the probabilities.
4. Probability of dependent events.
Suppose we are presented with 10 balls, five of which are black while the other five are white. What will be the probability of picking a black ball?
There are total of 10 balls and thus, there are 10 possible outcomes.
Now, let us denote the event of picking a black ball as A.
Hence, P(Black Ball) = P(A) = 5/10 = 0.5
Now comes the interesting part. Suppose you did pick a Black Ball at random. Now, you are left with 9 balls, 5 white and 4 black (because you have removed one Black Ball).
Now, if you had to pick one more ball from these 9, what would be the probability of picking one black ball?
P(Black Ball now) = 4/9 = 0.44
We can see that the probability changed in this case because the second event is actually dependent on the first event. This is called a dependent event.
Let's name the second event as event B. In this case, P(B|A) denotes the probability of the occurrence of B when A has already occurred.
Hence, P (second Black Ball) = P(B|A) = 0.44
In the above example, what would be the probability of selecting a Black Ball in the first try and selecting a Black Ball again in the second try?
We already know the probability of picking a Black Ball in the first try = P(red book) = P(A) = 5/10
and, P (second red book) = P(B|A) = 4/9
Now, since both the events must occur, this case requires a multiplication of the two probabilities.
Thus, P(A and B) = P(A) * P(B|A)
= 5/10 * 4/9 = 20/90
Note the formula: P(A and B) = P(A) * P(B|A)
Congratulations, we have completed theory of important lesson in Numeric Ability section, If you have any queries, post them in comments. Practice more and more to increase the chances of getting selected in exams. Best of luck friends.
