# Can't seem to understand probability (1 Viewer)

#### Snowflek

##### Active Member
Hey guys, so I'm just trying to understand probability but I just cant get my head around it... It just seems so illogical to me. Its like nonsense to me... I'm just reading through the examples of the exercise book' New Century Maths' but I'm still not understanding.... So my question is, how long did it take you guys to learn probabilty and how can I improve on it... #Frustrated...

#### BLIT2014

##### The pessimistic optimist.
Moderator
Post an example of the question you are struggling with and maybe one of us can "explain" the concept in a method that you will understand?

#### Snowflek

##### Active Member
Post an example of the question you are struggling with and maybe one of us can "explain" the concept in a method that you will understand?
1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...

#### InteGrand

##### Well-Known Member
1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...
For Q2:

a) Note that the number made will be greater than 500 if and only if the first digit is chosen to be either 5, 8, or 9. There are four choices for the first digit altogether (1, 5, 8 or 9), and precisely three of them result in a number larger than 500 (5, 8 or 9). Hence the answer is 3 in 4, i.e. 0.75 probability.

b) It is even if and only if the last digit is chosen to be 8. This happens with probably 1/4, since there are four choices for the last digit, and only one of them is desirable (8).

c) It is divisible by 3 if and only if the digit sum of the number is a multiple of 3. Note 5 + 9 + 8 + 1 = 23. So the only way to get the digit sum to be a multiple of 3 is if the digit we choose to leave out is 5 or 8. The probability of leaving out 5 is 1/4, and the probability of leaving out 8 is 1/4. Since the events "leave out 5" and "leave out 8" are mutually exclusive, the answer is Pr(leave out 5) + Pr(leave out 8) = 1/4 + 1/4 = 1/2.

#### InteGrand

##### Well-Known Member
1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...
$\bg_white \noindent The no. of total possible hands of five cards is \binom{52}{5}. The no. of hands that give a royal flush is just 4 (one way from each suit). Hence the probability of getting a royal flush is \frac{4}{\binom{52}{5}} = \frac{1}{649740}.$

$\bg_white \noindent An expression \binom{n}{k} is a \textsl{binomial coefficient}, and you can find out more about them here:$

https://mathlesstraveled.com/2007/02/14/binomial-coefficients/ .

#### ProdigyInspired

1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...
For 1, use npr to calculate how many combinations of 5 are in a 52 deck. Now, since a royal flush is defined as 10JQKA with the same suit, there are 4 combinations (4 suits). Therefore the probability is 4/Combinations 5 in 52. This will give you your answer.

#### sida1049

##### Well-Known Member
1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...
Probability trees are stupid. They work for some people, but they've never made sense to me. I'll offer a more abstract and general way of thinking about probability.

In general, the kind of probability you'll encounter is an equiprobable space, i.e. we're assuming that every outcome in the sample space (all possible outcomes) has the same probability/chance of happening. In this kind of probability, the chance of an event (which is a set of particular outcomes) happening is given by (the number of outcomes that event can happen) / (total number of possible outcomes). The numerator is the size of the event (the number of outcomes classified as that event), and the denominator is the size of the sample space (total number of outcomes). Remember, an event and the sample space are both sets of outcomes.

An example of this is what is the probability of picking an ace from a deck of cards? Our event is picking any ace. Our deck of cards is the sample space. The size of our event (picking an ace) is 4, because there are 4 aces in the deck, so picking any one of the aces is an outcome within our event. The size of our sample space is 52, because there are 52 possible outcomes.

So therefore, the probability of picking an ace from a deck of cards is (size of event)/(size of sample space) = 4/52 = 1/13.

1. Before I go through this problem in detail, please note that this is not an easy problem to learn probability with. In fact, this is the kind of problem 3 unit maths students study (permutations and combinations), so this would be a really tough problem for general maths. This problem uses the a technique that isn't obvious if it's the first time you've seen it. I will use and explain this technique in question 1 without referring to it, and name it in question 2.

It's tempting to say that our sample space is 52, since we have a deck of cards, but that would be incorrect. The question asks for 5 cards, so our sample space should be all possible ways of drawing 5 cards. We only need the size of the sample space. To find the size...

How many ways can we draw the first card? We have a deck of 52 cards, so there are 52 ways of drawing the first card.

How many ways can we draw the second card? Now that we only have 51 cards left in our deck, there are 51 cards of drawing the second card.

Repeating this reasoning, there are 50 ways of picking the third card, 49 ways for the fourth, and 48 ways for drawing the fifth.

Now how many ways can we draw 5 cards all together? The answer is 52×51×50×49×48. Why are we multiplying those numbers? I'll give you an explanation by considering an easier example: how many ways can be draw 2 cards? Suppose my first card is the ace of spades. Then there are 51 ways for me to choose the second card to get a unique pair of cards. Now suppose my first card is the ace of hearts. Then there are another 51 ways of drawing a unique pair of cards, so that means so far, we have 51+51=102 unique pairs of cards. Now if we do this for all 52 possibilities of drawing the first card, then we have 52×51 ways of drawing a pair of cards. So for 5 cards, we have 52×51×50×49×48 number of ways of drawing unique hands. (In general, and while this isn't exact, you can use the rule of thumb that when you see "and" in a probability question, you multiply, and when you see "or" in a probability question, you add.)

Now that we have the size of our sample space, we need the size of our event space. In a single suit, there is only one way of getting 10, J, Q, K and A in a random draw of 5 cards in that order. But in our case, the order does not matter, so we need to count the number of ways to draw those 5 cards in particular. But don't freak out, because we can solve this minor problem exactly like how we found the size of the sample space.

We have 5 unique cards. (the suit doesn't matter; we'll deal with it afterwards.) The number of ways we can draw the first card is 5. The number of ways we can draw the second card is 4. The number of ways we can pick the third card is 3, for the fourth card we have 2 ways, and the last card has just one way because we only have one card left. So we have 5×4×3×2×1=120 ways of rearranging 5 cards (or, if you like, we have 120 ways of drawing 5 cards from a deck of five; probability problems can be interpreted in many ways).

Remember there are 4 suits, so the size of our event is 4×120 = 480.

Now we have our size of event, and size of our sample space. So therefore, the probability of a royal flush is 480/(52×51×50×49×48)=1/649740. In other words, we have 1 in 649,740 chance of getting a royal flush.

2. Our sample space is all possible ways of making 3-digit numbers with those 4 digits. By the same technique that we have used in the previous question (called the multiplication principle), the number of ways to draw the 4 digits (or, if you prefer, the number of ways of drawing 3 cards from a deck of 4) is 4×3×2=24. Note that the multiplication principle takes into account ordering, which is relevant in this question. All we need to do for the following subproblems is to find the size of the events.

(a) As long as the 3-digit number doesn't start with 1, we're good. That means we need to find all possible 3-digit numbers with 5, 1, 9 and 8 such that the number doesn't start with 1. We're trying to fill _ _ _. How many numbers go into the first spot? 3, because we don't include 1. Second? 3, because we now have 3 numbers left, and remember that 1 can go into here. Last? 2. 3×3×2 = 18.

(b) The number is even as long as the last digit is 8. So we need to find the number of ways we can make numbers of the form _ _ 8 with the other digits. The problem is the same as the number of ways we can draw 2 cards from a deck of 3. So 3×2=6 number of ways. Another way to think about the problem is this: How many numbers can we put into the first digit? 3. How many for the second digit? We've used a digit already, so 2. Thus 3×2=6 ways.

So our answer is 6/24=1/4 again.

(c) A number is divisible by 3 if its digits add to 3. Thus we can have numbers with digits {5, 9, 1} or {9, 8, 1}. The number of ways of making a 3-digit number with 3 digits is 3×2×1=6 (again, it's the same as the number of ways of drawing 3 cards with a deck of 3 cards). Now we have 2 sets of 3 digits, so we multiple our previous answer by 2 to get the event size: 2×6=12.

Hope this helps!

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#### ProdigyInspired

I liked using probability trees as they gave me a graphical way to depict how many there were. However it only works for certain questions. Never used it again after Y11.

Moderator

#### Snowflek

##### Active Member
Probability trees are stupid. They work for some people, but they've never made sense to me. I'll offer a more abstract and general way of thinking about probability.

In general, the kind of probability you'll encounter is an equiprobable space, i.e. we're assuming that every outcome in the sample space (all possible outcomes) has the same probability/chance of happening. In this kind of probability, the chance of an event (which is a set of particular outcomes) happening is given by (the number of outcomes that event can happen) / (total number of possible outcomes). The numerator is the size of the event (the number of outcomes classified as that event), and the denominator is the size of the sample space (total number of outcomes). Remember, an event and the sample space are both sets of outcomes.

An example of this is what is the probability of picking an ace from a deck of cards? Our event is picking any ace. Our deck of cards is the sample space. The size of our event (picking an ace) is 4, because there are 4 aces in the deck, so picking any one of the aces is an outcome within our event. The size of our sample space is 52, because there are 52 possible outcomes.

So therefore, the probability of picking an ace from a deck of cards is (size of event)/(size of sample space) = 4/52 = 1/13.

1. Before I go through this problem in detail, please note that this is not an easy problem to learn probability with. In fact, this is the kind of problem 3 unit maths students study (permutations and combinations), so this would be a really tough problem for general maths. This problem uses the a technique that isn't obvious if it's the first time you've seen it. I will use and explain this technique in question 1 without referring to it, and name it in question 2.

It's tempting to say that our sample space is 52, since we have a deck of cards, but that would be incorrect. The question asks for 5 cards, so our sample space should be all possible ways of drawing 5 cards. We only need the size of the sample space. To find the size...

How many ways can we draw the first card? We have a deck of 52 cards, so there are 52 ways of drawing the first card.

How many ways can we draw the second card? Now that we only have 51 cards left in our deck, there are 51 cards of drawing the second card.

Repeating this reasoning, there are 50 ways of picking the third card, 49 ways for the fourth, and 48 ways for drawing the fifth.

Now how many ways can we draw 5 cards all together? The answer is 52×51×50×49×48. Why are we multiplying those numbers? I'll give you an explanation by considering an easier example: how many ways can be draw 2 cards? Suppose my first card is the ace of spades. Then there are 51 ways for me to choose the second card to get a unique pair of cards. Now suppose my first card is the ace of hearts. Then there are another 51 ways of drawing a unique pair of cards, so that means so far, we have 51+51=102 unique pairs of cards. Now if we do this for all 52 possibilities of drawing the first card, then we have 52×51 ways of drawing a pair of cards. So for 5 cards, we have 52×51×50×49×48 number of ways of drawing unique hands. (In general, and while this isn't exact, you can use the rule of thumb that when you see "and" in a probability question, you multiply, and when you see "or" in a probability question, you add.)

Now that we have the size of our sample space, we need the size of our event space. In a single suit, there is only one way of getting 10, J, Q, K and A in a random draw of 5 cards in that order. But in our case, the order does not matter, so we need to count the number of ways to draw those 5 cards in particular. But don't freak out, because we can solve this minor problem exactly like how we found the size of the sample space.

We have 5 unique cards. (the suit doesn't matter; we'll deal with it afterwards.) The number of ways we can draw the first card is 5. The number of ways we can draw the second card is 4. The number of ways we can pick the third card is 3, for the fourth card we have 2 ways, and the last card has just one way because we only have one card left. So we have 5×4×3×2×1=120 ways of rearranging 5 cards (or, if you like, we have 120 ways of drawing 5 cards from a deck of five; probability problems can be interpreted in many ways).

Remember there are 4 suits, so the size of our event is 4×120 = 480.

Now we have our size of event, and size of our sample space. So therefore, the probability of a royal flush is 480/(52×51×50×49×48)=1/649740. In other words, we have 1 in 649,740 chance of getting a royal flush.

2. Our sample space is all possible ways of making 3-digit numbers with those 4 digits. By the same technique that we have used in the previous question (called the multiplication principle), the number of ways to draw the 4 digits (or, if you prefer, the number of ways of drawing 3 cards from a deck of 4) is 4×3×2=24. Note that the multiplication principle takes into account ordering, which is relevant in this question. All we need to do for the following subproblems is to find the size of the events.

(a) As long as the 3-digit number doesn't start with 1, we're good. That means we need to find all possible 3-digit numbers with 5, 1, 9 and 8 such that the number doesn't start with 1. We're trying to fill _ _ _. How many numbers go into the first spot? 3, because we don't include 1. Second? 3, because we now have 3 numbers left, and remember that 1 can go into here. Last? 2. 3×3×2 = 18.

(b) The number is even as long as the last digit is 8. So we need to find the number of ways we can make numbers of the form _ _ 8 with the other digits. The problem is the same as the number of ways we can draw 2 cards from a deck of 3. So 3×2=6 number of ways. Another way to think about the problem is this: How many numbers can we put into the first digit? 3. How many for the second digit? We've used a digit already, so 2. Thus 3×2=6 ways.

So our answer is 6/24=1/4 again.

(c) A number is divisible by 3 if its digits add to 3. Thus we can have numbers with digits {5, 9, 1} or {9, 8, 1}. The number of ways of making a 3-digit number with 3 digits is 3×2×1=6 (again, it's the same as the number of ways of drawing 3 cards with a deck of 3 cards). Now we have 2 sets of 3 digits, so we multiple our previous answer by 2 to get the event size: 2×6=12.

Hope this helps!
Wow! Thank you very much for this massive explanation!! I appreciate it alot! I understand a lot more! Thank you!!!!! I'm very grateful!

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#### Snowflek

##### Active Member
For 1, use npr to calculate how many combinations of 5 are in a 52 deck. Now, since a royal flush is defined as 10JQKA with the same suit, there are 4 combinations (4 suits). Therefore the probability is 4/Combinations 5 in 52. This will give you your answer.
Thank you! This way seems so quick. Can you explain to me how to use nPr and nCr? I didn't get taught these in class and don't exactly know how to use them correctly.

#### Snowflek

##### Active Member
For Q2:

a) Note that the number made will be greater than 500 if and only if the first digit is chosen to be either 5, 8, or 9. There are four choices for the first digit altogether (1, 5, 8 or 9), and precisely three of them result in a number larger than 500 (5, 8 or 9). Hence the answer is 3 in 4, i.e. 0.75 probability.

b) It is even if and only if the last digit is chosen to be 8. This happens with probably 1/4, since there are four choices for the last digit, and only one of them is desirable (8).

c) It is divisible by 3 if and only if the digit sum of the number is a multiple of 3. Note 5 + 9 + 8 + 1 = 23. So the only way to get the digit sum to be a multiple of 3 is if the digit we choose to leave out is 5 or 8. The probability of leaving out 5 is 1/4, and the probability of leaving out 8 is 1/4. Since the events "leave out 5" and "leave out 8" are mutually exclusive, the answer is Pr(leave out 5) + Pr(leave out 8) = 1/4 + 1/4 = 1/2.
Thank you for the help!

#### ProdigyInspired

Thank you! This way seems so quick. Can you explain to me how to use nPr and nCr? I didn't get taught these in class and don't exactly know how to use them correctly.
Oh, my bad. I seemed to misunderstand that you were in 3U.

nPr and nCr are both functions on the calculator used in 3Unit, not sure if you can use it in General maths though as they may not teach you factorials and combination/permutation formulas (the name of the topic in 3U).

nPr represents permutations, where you would have

$\bg_white \frac{n!}{(n-k)!}$

Permutations represents the amount of possibilities a set of things (n) can be arranged (k) with replacement.
Meaning if you were to pick a 5 (n=52, k=5) card hand, you would pick one random card up, put it back, then pick another random one, then repeat.

nCr represents combinations, where you have

$\bg_white \binom{n}{k} = \frac{n!}{k!(n-k)!}$

This represents the amount of possibilities a set of things (n) can be arranged (k) without replacement.
Meaning if you were to pick a 5(n=52, k=5) card hand, you would pick one card, then proceed to pick another one without putting it back down.

On the calculator you can find them by pressing shift-multiply (npr) and shift-divide (ncr).

For example, if I wanted to find the total amount of possibilities of the above example, you would press 52, then nCr, then 5. This means my previous post was incorrect since its combinations not permutations (fixed).

Then to find the probability of a royal flush, you would do 4 / 52C5.

You can kind of say in your mind, [from] 52 I'll Choose 5, or for permutations from 52 I'll Pick 5.

Do NOT worry if you can't understand it. In 3U I'm pretty sure we spent at least a week going through the equation and deriving it to get an answer.

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#### BLIT2014

##### The pessimistic optimist.
Moderator
Oh, my bad. I seemed to misunderstand that you were in 3U.

nPr and nCr are both functions on the calculator used in 3Unit, not sure if you can use it in General maths though as they may not teach you factorials and combination/permutation formulas (the name of the topic in 3U).

nPr represents permutations, where you would have

$\bg_white \frac{n!}{(n-k)!}$

Permutations represents the amount of possibilities a set of things (n) can be arranged (k) with replacement.
Meaning if you were to pick a 5 (n=52, k=5) card hand, you would pick one random card up, put it back, then pick another random one, then repeat.

nCr represents combinations, where you have

$\bg_white \binom{n}{k} = \frac{n!}{k!(n-k)!}$

This represents the amount of possibilities a set of things (n) can be arranged (k) without replacement.
Meaning if you were to pick a 5(n=52, k=5) card hand, you would pick one card, then proceed to pick another one without putting it back down.

On the calculator you can find them by pressing shift-multiply (npr) and shift-divide (ncr).

For example, if I wanted to find the total amount of possibilities of the above example, you would press 52, then nCr, then 5. This means my previous post was incorrect since its combinations not permutations (fixed).

Then to find the probability of a royal flush, you would do 4 / 52C5.

You can kind of say in your mind, [from] 52 I'll Choose 5, or for permutations from 52 I'll Pick 5.

Do NOT worry if you can't understand it. In 3U I'm pretty sure we spent at least a week going through the equation and deriving it to get an answer.
You are allowed to use the nCr and nPr functions in general mathematics I used them for a fair few questions.

#### Snowflek

##### Active Member
Oh, my bad. I seemed to misunderstand that you were in 3U.

nPr and nCr are both functions on the calculator used in 3Unit, not sure if you can use it in General maths though as they may not teach you factorials and combination/permutation formulas (the name of the topic in 3U).

nPr represents permutations, where you would have

$\bg_white \frac{n!}{(n-k)!}$

Permutations represents the amount of possibilities a set of things (n) can be arranged (k) with replacement.
Meaning if you were to pick a 5 (n=52, k=5) card hand, you would pick one random card up, put it back, then pick another random one, then repeat.

nCr represents combinations, where you have

$\bg_white \binom{n}{k} = \frac{n!}{k!(n-k)!}$

This represents the amount of possibilities a set of things (n) can be arranged (k) without replacement.
Meaning if you were to pick a 5(n=52, k=5) card hand, you would pick one card, then proceed to pick another one without putting it back down.

On the calculator you can find them by pressing shift-multiply (npr) and shift-divide (ncr).

For example, if I wanted to find the total amount of possibilities of the above example, you would press 52, then nCr, then 5. This means my previous post was incorrect since its combinations not permutations (fixed).

Then to find the probability of a royal flush, you would do 4 / 52C5.

You can kind of say in your mind, [from] 52 I'll Choose 5, or for permutations from 52 I'll Pick 5.

Do NOT worry if you can't understand it. In 3U I'm pretty sure we spent at least a week going through the equation and deriving it to get an answer.
Thank you for taking the time to help me!! Thank you!!!