I found this post by Rahul an expert of Gurome. I am afraid Rahul isn't active, and would like to solicit an objective opinion of his colleagues at Gurome about my doubts.
How relevant would be to say the events are independent and consider the combination sets and not the single event or replicated events tied to certain probability(ies)? I have doubt in this post, as in selecting 4 students out 12 students (5 boys and 7 girls) the events cannot be independent. Thus the occurrence of one, e.g. selected student is boy will affect the chance of occurrence of the other, e.g. probability the student selected is boy (or next student is girl).
Please comment Gurome experts
Rahul@gurome wrote:The basic idea to solve this problem lies in the theory of mutual independence of events and mutual exclusiveness of events. And this concept is necessary and crucial not only for combinatorics but for probability too. The following two cases tries to explain the concepts.
Case 1: A group of students consists of 5 boys and 7 girls. In how many ways 4 students can be selected such that two of them are boys and other two are girls?
- In this case, the events are: selecting 2 boys out of 5 and selecting 2 girls out of 7.
Selection of boys and girls are mutually independent events. Because the events are occurring simultaneously but they do not affect each other. Thus they are combined events.
For a particular boy we can select any two girl out of 5 and vice versa.
Now number of ways to select two boys out of 5 = 5C2 and number of ways to select two girls out of 7 = 7C2.
Thus, number of ways 4 students can be selected such that two of them are boys and other two are girls = (7C2)*(5C2) = 21*10 = 210
Case 2: A group of students consists of 5 boys and 7 girls. In how many ways 4 students can be selected such that at least two of them are boys?
- In this case, the events are: selecting 2 or 3 or 4 boys out of 5.
These events are mutually exclusive events. Because the events are totally different and cannot occur at the same time. Thus they are separate events.
Already we know that selection of two boys will result in 210 combinations. Same procedure can be applied for 3 and 4 boys cases.
3 boys, 1 girl --> (5C3)*(7C1) = 10*7 = 70
4 boys, 0 girl --> (5C4)*(7C0) = 5*1 = 5
Total number of ways 4 students can be selected such that at least two of them are boys = (210 + 70 + 5) = 285
In short we can say,
- Events occur at the same time => Combined event --> Multiplication
Events do not occur at the same time => Separate event --> Addition
Hope it helps.
