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1) In a coin collection, each coin has some combination of the following characteristics:

One of five different colors (white, black, silver, gold, copper)

One of three different shapes (circle, square, hexagon)

One of three letters imprinted on it (A, B, C)

There is exactly one coin with each combination of characteristics. There is one black circle coin with an A on it, one gold square coin with a C on it, and so on.

a. How many coins are in this collection?

I think this is just the Fundamental Counting principle so it is just $5*3*3=45$

b. How many silver coins are in the collection?

I'm confused on this one..

There are $3*2=6$ total outcomes of shapes and letters. Do I just multiply that by $5C1$

c. How many coins have the letter A on them?

Again, I'm confused, is it: $3C1 * 5*3$?

2) Jeff and Caitlin are playing a game. Jeff chooses 4 balls from a bucket of 18 balls numbered 1 to 18. To win, Caitlin must correctly guess the numbers on the four balls.

a. How many ways can Jeff choose four balls?

$18C4$

b. What is the probability that Caitlin correctly guesses the four numbers?

I'm confused about this too. Is it $1/18*1/18*1/18*1/18$

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How many coins are in this collection?

Your answer is correct.

How many silver coins are in this collection?

We are told that there is one coin with each combination of characteristics. Since we are told that the coin is silver, there is only one way to choose its color. There are three choices for its shape and three choices for the imprinted letter. Hence, there are $1 \cdot 3 \cdot 3 = 9$ silver coins.

How many coins have the letter A on them?

Since we are told the coin has the letter A on it, there is only one way to choose its letter. There are five ways to choose its color and three ways to choose its shape. Hence, there are $5 \cdot 3 \cdot 1 = 15$ coins with the letter $A$ imprinted on them.

How many ways can Jeff choose four balls?

Your answer is correct.

What is the probability that Caitlin correctly guesses the four numbers?

Method 1: Caitlin must guess all four numbers correctly, which can be done in $\binom{4}{4} = 1$ way, while selecting four of the eighteen numbers. Hence, the probability that Caitlin guesses all four numbers correctly is

$$\frac{\dbinom{4}{4}}{\dbinom{18}{4}}$$

Method 2: Let's correct your attempt.

Caitlin chooses four different numbers. Therefore, she has $18$ ways to choose the first number, $17$ ways to choose the second number, $16$ ways to choose the third number, and $15$ ways to choose the fourth number.

Of these choices, she has four ways to choose one of the four numbers Jeff chose on her first attempt. If she successfully chose one of those four numbers, there are three ways for her to choose one of the other three numbers Jeff chose on her second attempt. If Caitlin's first two choices were successful, there are two ways for her to choose one of the other two numbers Jeff chose on her third attempt. If all three of her first three attempts were successful, there is one way for her to choose the remaining number Jeff chose on her fourth attempt.

Hence, the probability that Caitlin chooses all four numbers that Jeff selected is $$\frac{4}{18} \cdot \frac{3}{17} \cdot \frac{2}{16} \cdot \frac{1}{15}$$

You should verify that the two methods yield the same answer.

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There are $3×3=9$ silver coins . You can make a tree for good visualization .

There are 5 colors , 3 shapes branch and then there are 3 sub branch of letters on it .

Now for letter A put letters on top of tree so there are $5×3=15 $ Coins .

As for your next question Does Caitlin need also to guess the order of balls drawn . In that case answer is $${1\over18}×{1\over17}×{1\over16}×{1\over15}$$

If not then answer is simply $$\frac{1}{18\choose4}={1\over3060}$$[ 1 way to choose exact same balls]

Pretty low chances .

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  • $\begingroup$ Thanks , I edited my answer ! See if it is good now . $\endgroup$ – Rishi Mar 7 '20 at 13:03
  • $\begingroup$ The calculation $$\frac{1}{18} \times \frac{1}{17} \times \frac{1}{16} \times \frac{1}{15}$$ is still incorrect. To see why, please read the second method in my answer. $\endgroup$ – N. F. Taussig Mar 7 '20 at 14:54
  • $\begingroup$ I also added that if order of balls drawn matters $\endgroup$ – Rishi Mar 7 '20 at 15:32

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