Question 1

arent there 4! ways we can write the winning numbers . i mean the order doesnt matter so 3,15,46,49 should be the same as 15,3,46,49 but sal says that theres only one way of getting the correct lottery numbers why is that?

Accepted Answer

Achu and Naveen,
Sal could have solved this problem in two ways.  He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or—he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did).
Let us do it both ways, using the permutations first.
As you mentioned, there a 4! ways of writing the four numbers.  Another way to say this is that there are 4! different ways to order the four numbers –or-- there are 4! different permutations of the four numbers that give us the favorable outcome.  This can be written as 4*3*2*1.
The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57.
Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57).  This simplifies to  1/487,635.
Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu).
The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!).  This can be written as (60*59*58*57)/(4*3*2*1) .  This number is 487,635.
Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635.  This is the same answer we got using permutations.
Consider combinations and permutations to be different “units”.  You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½.  You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability.   Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability.

Question 2

Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic:

If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin with
4/60 as my chances of getting one of those numbers on that first draw.

On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply:
(4/60) * (3/59)

The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is:

(4/60) * (3/59) * (2/58) * (1/57)

I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to

4! / (60! / (60-4)!) 
or
4! / P(60,4)

I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60.

This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. Can anyone do a better job of explaining why my intuition was right?

Accepted Answer

This sounds like a tautology but your intuition is right because it is right.
Let's look at the different ways of arriving to the answer.

Your way: look at the situation one draw at a time.
4/60 * 3/59 * 2/58 * 1/57, which is equal to 4*3*2*1 / (60*59*58*57)

Sal's way:
4! / (60 P 4)
which is the same as
4! / (60!/(60-4)!)
which is the same as
4! * (60-4)! / 60!
which is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)
So your and Sal's ways are the same thing just expressed in different ways.

Here's one more way to look at it:
There are (60 C 4) ways of choosing four numbers (all different) from 60. Only one combination of them is what we have chosen. So the probability is
1/(60 C 4)
which is the same as
1/( 60!/(4!*(60-4)!))
which is the same as
4! * (60-4)! / 60!
which, as we already have seen, is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)

Question 3

If S=1+2+4+8+16+32................ Can this be taken as S=1+2(1+2+4+8+16.......)???
If it can be taken,
                       S=1+2S
                       S=-1      Is this a valid answer for such a big question???

Accepted Answer

Your reasoning only works when the sum S is a real number and does not continue on to infinity.  For example, if S = 1 + 1/2 + 1/4 + ... + 1/(2^n) + ... and so on forever, then your logic says that S = 1 + 1/2(S), which gives the right answer of S = 2.

The difference between the two problems is that the sum S = 1 + 2 + 4 + ... continues to get larger forever and does not stop, while S = 1 + 1/2 + 1/4 + ... has a limit that it can never reach, no matter how long you count.  By taking your approach, you can short-cut the infinite series and figure out that limit without having to take forever (literally!).

Question 4

Shouldn't the probability of correctly choosing the four numbers be 1/60 * 1/59 * 1/58 * 1/57? Why would you need to use choose in this case? Don't we have a 1/60 chance of getting the 1st number, 1/59, and so on?

Accepted Answer

The expression 1/60 * 1/59 * 1/58 * 1/57 represents the probability that the first winning number chosen is 3, the second winning number chosen is 15, the third winning number chosen is 46, and the fourth winning number chosen is 49.  So your method specifies an order.

However, in the problem asked, the order of the winning numbers does not matter.  The problem asks for the probability that the first, second, third, and fourth winning numbers are 3, 15, 46, 49 in any order.  This is why "choose" is used.

If you want to solve this problem using selection without replacement, two possible equivalent methods are
1/60 * 1/59 * 1/58 * 1/57 * 4! and
4/60 * 3/59 * 2/58 * 1/57.

Have a blessed, wonderful day!

Question 5

if in this lottery, picking a number and putting it back is allowed... so that means you can pick a number a multiple of times... what would the probability be then?
64! / ((64-4!) * 4! )   ?

Accepted Answer

Well, you'd choose 4 numbers from 60 numbers (1 to 60) and repetition is allowed, the probability of winning would be 1/(60^4/4!) = 4!/60^4 = 1/540000 ≈ 0.000002
(The 4! is there because the order of the numbers still doesn't matter. If the winning row was 7,34,34,50 and you had 34,7,50,34 you'd still win)

Question 6

I think I may have a fundamental misunderstanding of combinations and / or permutations. I tried to solve this problem by doing the following (60! / (56! * 4!)) / (60^4) which is the combinations formula divided by (I thought) the total number of possible outcomes with 60 numbers in 4 slots. Why is that incorrect?

Accepted Answer

60^4 isn't the total number of possible groups of 4, because the order of the 4 numbers doesn't matter for combinations. 60^4 is the number of permutations, not combinations.
and 60!/(56!*4!) is the total number of possible combinations of 4 numbers, so it is the sample space, not the # of successes meaning it should be the denominator.

Question 7

I was just wondering what the nCr and nPr buttons on the calculator do. My teacher explaned it, but i forgot what the do and how to use them. Please help!

Accepted Answer

nCr is used for Combinations, while nPr is used in permutations. 'N' represents the total number of items you have to choose from, and 'R' represents the number you choose. So if you had 36C10, that would mean you have 36 items and you can choose 10, regardless of order, since it is a Combination.

Question 8

Isn't 59 factorial (!), 60*59*58*57*56*all the way down to 0??

Accepted Answer

No, there's no 60 or 0 involved. It's 59 through 1.

Question 9

In the same Sal's case - 4 number ticket, let's assume that if I pay a little bit more for the lottery ticket I'm able to pick 5 numbers instead of 4. Then I would have a higher probability of winning. But how do I calculate that probability with an "extra number"?

Accepted Answer

Let find the easy part, which is the total # of outcomes. This is just going to be the number of different ways in which the host of the lottery prize is going to choose 4 numbers out of 60. This is just C(60,4).

Now the # of ways in which our event can happen is just the number of ways in which those 4 numbers match exactly 4 of our 5 numbers. i.e. 4 out 5 numbers: C(5,4).

Thus the probability is:

5/487,635 = 1/97,527

And it makes sense. If you paid the number 1, 2, 3 ,4 and 5, you will win in these situations:

[1 - 2 - 3 - 4]
[1 - 2 - 3 - 5]
[1 - 2 - 4 -5]
[1 - 3 - 4 -5]
[2 - 3 - 4 -5]

Course: Precalculus > Unit 8

Example: Lottery probability

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