5th Combinatorial question

[pivoxa15]

Homework Statement

To win Division 1 in the game of Tattslotto, the player must have the same 6 numbers (in any order) as those that are randomly drawn from the numbers 1 to 45. A Division 2 prize requires that the player’s ticket must have 5 of the 6 winning numbers and a Division 3 prize requires that the player has 4 of the 6 numbers drawn. Calculate the probability that the player’s 6 numbers will contain at least a Division 3 prize.

Homework Equations

Pr(event) = Favourable possibilities/total possibilities

C means ‘choose’ i.e. nCr=n!/(r!(n-r)!)

The Attempt at a Solution

Assume player has chosen their 6 numbers and the officials have drawn the 6 numbers.

Favourable possibilities = {6C6=1, 6C5=6, 6C4=15}
Total possibilities = 45C6

Pr(event) = (1+6+15)/45C6 = 22/8145060

But the answers suggested 1135/8145060

 

[neurocomp2003]

have you learned the inclusion /exclusion principle...if not i suggest looking it up.

If you can't find it then think about what are the relations between
6C6 6C5 6C4

also your summation is wrong...reread the question particularly the last 2 statements.

 

[Dick]

Funny. I get 1135/814506. Your stated answer is not in lowest terms, so I suspect a typo. You are only accounting for the ways of choosing winning numbers, you have to choose the correct number of losing numbers in each case as well.

 

[pivoxa15]

have you learned the inclusion /exclusion principle...if not i suggest looking it up.

If you can't find it then think about what are the relations between
6C6 6C5 6C4

also your summation is wrong...reread the question particularly the last 2 statements.

I am a bit lost here. I understand the inclusion/exclusion principle but don't see how it's applied here.

 

[Dick]

I am a bit lost here. I understand the inclusion/exclusion principle but don't see how it's applied here.

I don't either, but there are many more ways of picking four winning numbers than 6C4. You have two more numbers to choose.

 

[neurocomp2003]

my bad read the question wrong.

 

[pivoxa15]

I don't either, but there are many more ways of picking four winning numbers than 6C4. You have two more numbers to choose.

Right. To get either 1st, 2nd or 3rd prize, the total number of combinations is 6C6*39C0 + 6C5*39C1 + 6C4*39C2. Total possibilities is 45C6.

So its (1+234+11115)/8145060 = 1135/814506