# Lesson: The Complement

## Comment on The Complement

### In the "odd product of three

In the "odd product of three numbers" question, you used the fundamental counting principle. Could you use combination instead, e.g. "2C1, 5C1, 4C1"? Obviously those are still 2x5x4, but I wanted to make sure that wasn't just--well--lucky!

I ask mostly because, as I think of the steps in getting to 40, it seems--at least to me--that order does not matter, and thus combination could(?) be used.

Thank you!

### That's a totally acceptable

That's a totally acceptable approach. It just turns out that nC1 always equals n (for positive integer n)

### I solved it this way-

I solved it this way-

P(Product of three numbers) = 1/2*1/5*1/4 = 1/40
P(Product of three numbers is even) = 1- P(product is not even) = 1- P(product of three numbers)= 1-1/40 = 39/40. Is this approach correct?

### Hi Deepak,

Hi Deepak,

I'm not sure what you mean by "P(Product of three numbers) = 1/2*1/5*1/4 = 1/40"

Did you mean to say "P(Product of three numbers IS ODD) = 1/2*1/5*1/4 = 1/40"??
If that's the case, then your approach is perfect.

Thanks Brent!

### I have one more doubt

I have one more doubt regarding the second example of probability of product of 3 numbers which will be even.
It is about the probability formula:
P(Product odd)= no. of outcomes where the product is odd/total number of possible outcomes.

shouldn't the denominator be total number of possible products? This doubt may seem silly, but it just came to my mind that why is the denominator not total number of possible products. Please clarify.

### Question: Shouldn't the

Question: Shouldn't the denominator be total number of possible products?

Answer: No, we need to consider all possible outcomes.

Consider the following question:

Bag A contains one hundred balls. Ninety-nine of the balls have a 2 written on them, and one ball has a 1 written on it.

Bag B also contains one hundred balls. Ninety-nine of the balls have a 2 written on them, and one ball has a 1 written on it.

If one ball is randomly selected from each bag, what is the probability that the product of the two balls is 1?

As one might guess, P(product is 1) is VERY UNLIKELY, since we need to select the one 1 from each bag.

However, if we only examine the possible PRODUCTS (as you suggest), we see that there are only 3 possible products: 1, 2 and 4
However, we CANNOT conclude that P(product is 1) = 1/3

Instead, we must examine all of the possible OUTCOMES.
There are 100 ways to select a ball from Bag A, and there are 100 ways to select a ball from Bag B.
So, the total number of OUTCOMES = 100 x 100 = 10,000

There is only ONE way to get a product of 1. We must select the one 1 from Bag A, and we must select the one 1 from Bag B.

So, P(product is 1) = 1/10,000

### Can the below question be

Can the below question be solved using complement as the word at least is used?-
The question:

Throw: ----- Outcome:
-- 1 ---------- Lose \$3.00
-- 2 ---------- Lose \$2.00
-- 3 ---------- Lose \$1.00
-- 4 ---------- No Effect
-- 5 ---------- Win \$1.00
-- 6 ---------- Win \$5.00

The results of throwing a single die in a certain gambling game are shown in the table above. What is the probability that a player will have won at least \$5.00 after two throws?

A) 1/36, B) 1/12, C) 1/9, D) 5/36, E) 1/6

### The complement would take

The complement would take wayyy too long. Here's what I'd do:

We'll have to assume that rolling each outcome (1, 2, 3, 4, 5, or 6) is EQUALLY likely (1/6 for each).

There are 36 possible outcomes in total (6 outcomes for the first roll, and 6 outcomes for the second roll.)

Let's count the number of ways to win AT LEAST \$5.00
- 6 on 1st roll (win \$5), and 5 on 2nd roll (win \$1)
- 5 on 1st roll (win \$1), and 6 on 2nd roll (win \$5)
- 4 on 1st roll (no effect), and 6 on 2nd roll (win \$5)
- 6 on 1st roll (win \$5), and 4 on 2nd roll (no effect)
- 6 on 1st roll (win \$5), and 6 on 2nd roll (win \$5)
So, there are 5 possible ways to win AT LEAST \$5

So, P(win at least \$5) = 5/36

Cheers,
Brent