Niha01 wrote:
HongHu wrote:
Good job thearch. You can also do it this way: 2^5-5(four wins)-1(five wins)=26.
Can you please elaborate on this approach
Posted from my mobile deviceHi Niha01,
Five hands in general, could have any of the following outcomes:
Case 1) All 5 wins
Case 2) 4 wins and 1 loss
Case 3) 3 wins and 2 losses
Case 4) 2 wins and 3 losses
Case 5) 1 win and 4 losses
Case 6) All 5 losses
In the given question, we know there are 3 wins and 9 losses. Hence, the first five hands cannot have more than three wins. Hence cases 1) and 2) are not possible while any of the cases 3), 4), 5) or 6) are possible.
Now, there are two approaches to solve this question.
1st Approach:
Count the outcomes of the cases that are possible which is:
Case 3) 5C3 = 10
Case 4) 5C2 = 10
Case 5) 5C1 = 5
Case 6) 5C0 = 1
So, total possible outcomes = 10 + 10 + 5 + 1 = 26
2nd Approach: Subtract the outcomes of the cases that are not possible from the total no. of outcomes.
Case 1) 5C5 = 1
Case 2) 5C4 = 5
These two cases have to be subtracted from total no. of cases.
But how to find total no. of cases?
Each of the five hands has 2 possibilities - win or loss.
So five hands will have 2 x 2 x 2 x 2 x 2 = 32 possible outcomes.
So we subtract cases 1) and 2) from the total i.e. 32 - 1 - 5 = 26.
I hope this is understood. (Kudos if you have.)
Choice of which approach to use depends on which calculation will be lesser. If cases to be counted are less, use 1st approach. If cases to be excluded are less, use 2nd approach.
Hope this helped.