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# Probability

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66
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In Ms. Q's deck of cards, every card is one of four colors (red, green, blue, and yellow), and is labeled with one of seven numbers (1, 2, 3, 4, 5, 6, and 7). Among all the cards of each color, there is exactly one card labeled with each number. The cards in Ms. Q's deck are shown below.

Yunseol draws 5 cards from Ms. Q's deck. What is the probability that three cards have the same number?

Dec 31, 2022

#1
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Whoa, this problem looks a bit tricky...

red

green

blue

yellow

And each card has a number:

1

2

3

4

5

6

7

We also know that we should be finding probability which would be in the form of:

successes/total.

We should start by finding the total number of cards:

There are 4 options for the colors, and 7 options for the numbers so:

4 * 7

= 28 cards in the deck

Then, we are going to be drawing 5, and of the 5, 3 of them have to include the same number.

We can pick the number that all three have in 7 ways,

Then, we can pick choose 3 of the 4 for our different colors,

So, we have

7 * C(4,3) = ?

7 * 4 = ?

28 ways, now we have to pick the other two cards...

Since there are going to be 25 cards left, we have

28 * 25

Then since we are going to be picking another one out of the 24 that are left:

28 * 25 * 24

Now let's find the total number of ways that you can draw 5 cards:

28 * 27 * 26 * 25 * 24

So our probability is:

(28 * 25 * 24)/ (28 * 27 * 26 * 25 * 24)

Simplifying we get:

1/(27 * 26)

= 1/702

So the probability is 1/702 that you get the same number 3 times out of 5.

(If anyone can back me up or reject my answer completely that would be great. I also advise you to read my answer, because I might have made a mistake somewhere ;)

Dec 31, 2022
#3
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I believe I did say: "might have made a mistake somewhere, why don't you look carefully and read it, instead of just going after the answer"

TooEasy  Jan 2, 2023
#4
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Posting alibi disclaimers to negate and shirk responsibility for mathematical SLOP

are hallmarks for mathematicians who are worth less than nothing.

Solution:  My solution is Wrong and it’s SLOP...

Description:

Choose one (1) of the seven (7) numbers, then choose three (3) of the four (4) colors for these numbers.

Then choose one (1) of six (6) numbers and then choose one (1) of the four (4) colors for this number.

Then choose one (1) of six (6) numbers (again), and then choose one (1) of the four (4) colors for this number (again).

Then divide this by the $$\dbinom {27}{5}$$ways to choose a set of five (5) card from a population of 27 cards.

Descriptive Math:

$$\text {Ignore this slop}\\ \dfrac { \dbinom {7}{1} \dbinom {4}{3} \dbinom {6}{1} \dbinom {4}{1} \dbinom {6}{1} \dbinom {4}{1}} {\dbinom {27}{5}} = \underbrace {\left (\large \dfrac {896}{4485}\right)}_{\text {exact probability}} \left ( \approx 19.98\% \right )$$

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Even if I corrected the divisor, it’s still SLOP...

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To solve at this level requires hypergeometric distributive counting. (Tiggsy’s solution)

OR

Case analysis of binomial distribution. (Below)

$$\huge \rho(x) \normalsize = \dfrac { \left [\dbinom {7}{1} \dbinom {4}{3} \right] * \left[ \left (\dbinom {6}{2} \dbinom {4}{1}\dbinom {4}{1} \right ) + \left(\dbinom {6}{1} \dbinom {4}{2}\right) \right]} {\dbinom {28}{5}} = \underbrace {\left (\large \dfrac {46}{585}\right)}_{\text {exact probability}} \left ( \approx 7.86\% \right )\\$$

Expansion, Dissection, and Description:

nCr(7,1)*nCr(4, 3) * nCr(6,2)*nCr(4, 1)*nCr(4, 1) =  6720  |Counts of triples with non-pairs.

Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose two (2) of the six (6) remaining numbers; choose one (1) of four colors for each of the two (2) numbers.

nCr(7,1)*nCr(4, 3) * (nCr(6,1)*nCr(4, 2) = 1008 |Counts of triples and only pairs.

Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose one (1) of the six (6) remaining numbers; choose two (2) of the four (4) colors –making a pair of numbers.

Add these counts: 6720 + 1008 = 7728

Divide: 7728/ nCr(28,5) =  46/585  = 7.86%

GA

--. .-

GingerAle  Jan 2, 2023
edited by GingerAle  Jan 4, 2023
#5
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Here are my attempts, comments/refutations welcome.

First, a counting type approach.

The pack consists of 28 cards, (4 card colours and 7 numbers for each colour).

The number of 5 card combinations that can be drawn will be C(28,5).

Needed are 3 cards with the same number, so 7*C(4,3) = 28 possibilities.

Assuming that the possibility of having 4 cards with the same number is excluded, the remaining 2 cards have to come from 24 remaining cards, (4 cards each of the other 6 numbers), C(24,2) possibilities.

That generates a probability of     $$\displaystyle \frac{28*C(24,2)}{C(28,5)}=\frac{46}{585} \approx 7.86 \%.$$

Second, a probability type approach.

Choose a number and suppose that we draw, in succession, three cards with that number, followed by any two cards with a different number.

The probability of that sequence will be

$$\displaystyle \frac{4}{28}\times \frac{3}{27} \times \frac{2}{26}\times \frac{24}{25}\times \frac{23}{24} =\frac{23}{20475}$$

Now there are 7 different numbers to choose from and the sequence can occur in 10 different ways, so the final probability will be

$$\displaystyle 7 \times 10 \times \frac {23}{20475}= \frac{46}{585} \approx 7.86 \%, \text{as earlier}.$$

Jan 3, 2023
#6
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Hello Tiggsy:

Your math and descriptive solutions for both of your methods are clear and concise.

You counting method is an optimal presentation of hypergeometric distribution counting.

Wonderful!

Thank you,

GA

--. .-

GingerAle  Jan 4, 2023
#7
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Thankyou for the feedback GA, ..... and I shall endeavour to find out what hypergeometric distribution counting is.

Tiggsy

Tiggsy  Jan 5, 2023