# Math Help - Two Digits

1. ## Two Digits

10 cards with one integer between 1 and 5 written on them are inserted into a bag. Each integer will be written on two cards each. Two cards are taken from the bag, the first one takes the tens place, and the second one takes the ones place to create a two-digit-number.

If the probability for any card to be taken out of the bag is equal, answer the following;

(i) What is the probability for the two digits to be different integers?
(ii) What is the probability for the number to be less than 20?
(iii) What is the probability for the number to be a multiple of three?

I'm going to be completely honest here and say I have no clue how to even get started on this one >_<

Any input greatly appreciated.

2. the probability that the first digit is 1 is

$\frac{2}{10}$

the probability that the second one also is 1 is

$\frac{2-1}{10-1}$

since you can consider that one card is chosen a second or so after the first.

Since the probability is the same for the other 4 digits,
the probability of the two cards being the same is

$5\ \frac{2}{10}\ \frac{1}{9}=\frac{1}{9}$

3. For the 2-digit number to be <20,
the first digit can only be 1.

The numbers are

11, 12, 13, 14, 15

However, there are 2 of each number.

11 can occur in 2 ways
12 can occur in 4 ways if you imagine giving the digits different colours.

There are 18 possibilities from 10(9) in total.

Or, 5 possible indistinguishable 2-digit numbers from 5(5)=25 different possibilities.

4. The numbers range from 11 to 55.

The multiples of 3 in this range are
12, 15, 18, 21.......up to 54.

Any number ending in a digit >5 cannot appear,
hence

12, 15, 21, 24, 33, 42, 45, 51 and 54 are the possible numbers divisible by 3.

That's 9 out of 25

5. Hello, davidman!

10 cards with an integer from 1 and 5 written on them are inserted into a bag.
Each integer will be written on two cards each.
Two cards are taken from the bag (without replacement).
The first takes the ten's place, and the second takes the one's place to create a two-digit-number.

If the probability for any card to be taken out of the bag is equal, answer the following;

(a) What is the probability for the two digits to be different integers?
(b) What is the probability for the number to be less than 20?
(c) What is the probability for the number to be a multiple of three?

I have no clue how to even get started on this one . . . . Really?
I don't suppose you thought of writing out the possible outcomes . . .

. . . . . $\begin{array}{ccccc}11 & 12 & 13 & 14 & 15 \\
21 & 22 & 23 & 24 & 25 \\ 31& 32 & 33 & 34 & 35 \\ 41 & 42 & 43 & 44 & 45 \\
51 &52 & 53 & 54 & 55 \end{array}$

Can you answer the questions now?

6. P.S. davidman,

the probability of the first 2 digits being different is

$1-\frac{1}{9}=\frac{8}{9}$

7. I have resisted responding to this because of what I see a some basic confusions.
This is not a dice type problem.
The probability of getting $(n,n)$ is $\frac{2}{10}\frac{1}{9}=\frac{1}{45}$.

Whereas, the probability of getting $(m,n),~m\ne n$ is $\frac{2}{10}\frac{2}{9}=\frac{2}{45}$.

Thus the probability of getting a number less than 20 does involve the pairs
$(1,1),~(1,2), ~(1,3), ~(1,4), ~(1,5),$.

That is: $\frac{1}{45}$ $+ (4)\frac{2}{45}$

8. Thanks for all the help. I'm not quite sure how to get to some of these answers myself, though.

I guess I understand the probability of the first card you draw being a certain number.

$\frac{2}{10}=\frac{1}{5}$

since there's one less of the card, the probability goes down

$\frac{2-1}{10-1}=\frac{1}{9}$

the probability of getting the same number card both times is then

$5\times\frac{1}{5}\times\frac{1}{9}=\frac{1}{9}$

(why multiplied by 5?)

for the two cards to be different, though...

$1-\frac{1}{9}=\frac{8}{9}$

why is this? (this is actually the answer in my book)

This does unfortunately not add up with Plato's answer, which is the one I thought I understood the best...

As for a number less than 20, I can see how that becomes $\frac{5}{25}=\frac{1}{5}$

either by looking at the table of possible outcomes, or by adding the probabilities of "two same digits" and four outcomes of "two different digits".

I'm sure my translation of this problem is part of the actual problem here, but what I guess I want to know is why there are two versions here of the probability of getting two different digits.

9. Hi davidman,

we multiply by 5 because the calculation found only the probability
of a single pair of numbers being the same,
such as 11 or 22 or 33 or 44 or 55.

Hence the probability of 11 is \frac{1}{45}

the probability of 22 is \frac{1}{45}

the probability of 33 is \frac{1}{45}

the probability of 44 is \frac{1}{45}

and the probability of 55 is \frac{1}{45}

that's why we multiply by 5.

In calculating the probability of the 2 cards being different,
it's simpler to calculate the probability of them being the same first sometimes. Then, as all probabilities for a situation sum to 1, we can subtract that answer from 1 to get the answer you are looking for.

10. Thank you! It all makes so much sense now.