A doubt.

I came across this problem today,\[\text{Two standard die are thrown simultaneously,}\\ \text{what is the probability that the sum of the numbers they show is 9?}\] \[\text{Solution}: \text{First method}:\text{this is the usual method that is,consider}\\ \text{the ordered pairs.}\\ \text{Total number of ordered pairs satisfying the given}\\ \text{conditions is 4 viz. (3,6),(4,5),(5,4),(6,3).}\\ \text{Total number of ordered pairs=36}.\\ \text{Thus,P(E)}=\dfrac{4}{36}=\dfrac{1}{9}.\]
$\text{Second method}:\text{In this method we consider different}\\ \text{sums which are 2-12 that is a total of 11 sums.But only}\\ \text{one satisfies our condition which is 9.}\\ \text{Thus,P(E)}=\dfrac{1}{11}$ .Please help.

#Combinatorics #Probability

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Azhaghu Roopesh M @Satvik Golechha

Adarsh Kumar - 6 years, 3 months ago

The probability of appearing of the different sums from 2 to 12 is different because for 2, you have only 1 option, that is , (1,1), but for 5, you can have (1,4), (2,3), (3,2), and (4,1). So, method 1 is correct.

Satvik Golechha - 6 years, 3 months ago

oh!thanxxx!!

Adarsh Kumar - 6 years, 3 months ago

I'm truly very sorry dude , I didn't check my mail account today so I didn't come to know that you had mentioned me here . It's just by accident (I was checking the Explore tab for the first time !!) that I came to know of this .

Well, it seems Satvik has clarified your doubt :) . Sorry again !!

BTW I'm asking Calvin sir to devise a method so that we can get Notifications whenever someone mentions us in the Notifications tab . If possible reply to my comment at Calvin sir's messageboard so that he notices this inconvenience .

A Former Brilliant Member - 6 years, 3 months ago

Hey!No problem man!

Adarsh Kumar - 6 years, 3 months ago

My take on this is $\frac {2}{21}$ . When you throw a pair of dice, there are 21 possible outcome: 1-1,1-2 the same as 2-1, 1-3 the same as 3-1...and so on. That is equal to $_6C_2 + 6$ . Since there are 2 possible combinations that give a sum of 9: 6 and 3 combination and the 4 and 5 then it is equal $\frac {2}{21}$ .

Edit: This is under the assumption that one of the dice cannot be distinguished from the other. However, if one of the dice is colored blue then your first solution applies.

Roman Frago - 6 years, 3 months ago

You're making the same mistake as the OP. Just because there are 21 possible outcomes, doesn't mean they occur with the same probability. Your edit clearly shows why. How does changing the color of the die change the sum of the dice? What if I'm colorblind, does that mean that if someone watches the same dice rolls with me, he'll see different sums with different probabilities?

Take a simpler example of 2 dice. The possible outcomes according to you are (1,1),(1,2),(2,2).But if you try this example in real life, you'll find that (1,1) occurs half the times (1,2) does. Why? Because if we take ordered pairs, we can see that (1,2) can be achieved two ways, (1,2) and (2,1) whereas (1,1) can only be achieved one way (1,1).

Edit:- Just remembered an even simpler example. The lottery. Technicalities aside, there are only two outcomes, win or lose. But we all know that the probability of winning is not $\frac{1}{2}$

Siddhartha Srivastava - 6 years, 3 months ago

Thanxx!I got it!

Adarsh Kumar - 6 years, 3 months ago

What do you mean by "OP" ?

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Original poster.

Siddhartha Srivastava - 6 years, 3 months ago

Even if you are color blind that does not change the fact that one die can be distinguished from the other.

If the two dice can be distinguished from one another, they become independent events.

Roman Frago - 6 years, 3 months ago

@Roman Frago –

Even if you are color blind that does not change the fact that one die can be distinguished from the other.

Sorry. I thought your definition of being distinguishable was based on perception. Nvm. Anyways. So what you're saying means that being "distinguishable" is an intrinsic property of a dice. That is, It doesn't matter if I can't distinguish between them or not, they are intrinsically "distinguishable". And a pair of dice which does not satisfy this is "indistinguishable".

But this leads to absurd conclusions. Like for example, if I make a small dot on one of a pair of "indistinguishable dice" , the probability of getting a sum of 9 suddenly changes, even though the dot has negligible effect of the physical properties of the dice.

Or what about if I track one of the die with a laser pointer or with my eye? Even that makes the die distinguishable. This will lead to different outcomes just if I look at the die. I know this occurs in quantum mechanics, but dice are too big for it too apply.

Anyways, that was not even the main point I was trying to make. The main point was that the fact that there are 21 different outcomes does not mean the all occur with the same probability.

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – If the rolling of the dice is taken as a single event, then they have equal probability based on that condition. It is different from rolling one dice at a time. It also different if the question is asked after one of the dice has stopped rolling. Probabiblity changes depending on the condition.

Roman Frago - 6 years, 3 months ago

@Roman Frago –

If the rolling of the dice is taken as a single event, then they have equal probability based on that condition

Could you elaborate? I addressed this in my first comment. By the same logic, if I take winning the lottery as one event and not winning the lottery as the second, I have a $\frac{1}{2}$ chance of winning the lottery. Just because you define some things as events doesn't make the probability of them occurring equal.

Probabiblity changes depending on the condition.

I agree, but that doesn't mean that EVERY change in conditions will result in a change in probability. You are taking extreme examples in which either we have new information(what is on the first dice after the first dice has stopped rolling) or the whole experiment is changed ( rolling one die instead of 2.).

You also haven't answered what I wrote before. How does adding a small dot, or even just looking at a die change the probability of getting a certain sum.

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – The lottery analogy does not apply. It will if we say in a game dice, you win if you get a sum of 9.

In a lottery what are the possible outcome, not win or lose. The outcome is the possible combination of numbers. Winning or losing is the effect of the outcome.

Probability changes depending on the condition. Say you ask what is the probability of getting a sum of 9 when one of the dice has stopped rolling and showing one.

Roman Frago - 6 years, 3 months ago

@Roman Frago – In a lottery, what is the chance of getting 1,2,3,4,5,6? Is it different from getting 6,5,4,3,2,1?

Roman Frago - 6 years, 3 months ago

@Siddhartha Srivastava – A small dot makes the dice distinguishable which makes the condition totally different. They can't be taken as a single event but 2 events.

Addendum: With distinguishable dice, you can identify 2-1 from 1-2, even 1-1 from 1-1 making them additional outcome.

Roman Frago - 6 years, 3 months ago

The analogy does not apply...

Roman Frago - 6 years, 3 months ago

LOL this was a question in my model exam which I badly flunked >:o

Krishna Ar - 6 years, 3 months ago

BTW Which model exam , do you mean for Boards ?

A Former Brilliant Member - 6 years, 3 months ago

Yup! :P

Krishna Ar - 6 years, 3 months ago

@Krishna Ar – Aren't you in CBSE? They still have the CGPA system right? Don't Worry. Everyone gets CGPA 10. Everyone.

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – Agreed !! 2 years ago there were 71 students who got 10 CGPA !!

If you want , you can visit this link to see the names :)

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – I think you're confused. That is the list of poeple who got CGPA from ONE school.A lot of people get CGPA 10. In the order of thousands. Most schools usually have 10 - 20 students getting CGPA 10 at worst. My school had 30 students and another school had 150.

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – Sorry bro , I forgot to mention that the list I that I had provided was of my school . Yes , I was actually giving an example to prove your point that CBSE Boards are scoring:)

A Former Brilliant Member - 6 years, 3 months ago

@Siddhartha Srivastava – Yup. That's true. That's mainly because of the upgradation system. If that weren't there, the number would be halved or reduced to a third.

Krishna Ar - 6 years, 3 months ago

@Krishna Ar – Hey Krishna , I wanted to ask this for some time now , is Krishna Ar your real name or is it an alias ?

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Alias. Not too keen on revealing my name(at least as of now)

Krishna Ar - 6 years, 3 months ago

@Krishna Ar – Fine with me ,I was just curious :)

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – haha :)

Krishna Ar - 6 years, 3 months ago

@Krishna Ar – I know, I benefited from the same. :D

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – Haha , now my parents know about this, so they want a cgpa 10 w/o upgrades. (in fact, they want centums too :3)

Krishna Ar - 6 years, 3 months ago

@Krishna Ar –

(in fact, they want centums too :3)

I thought they only tell you your grades? Not your marks. I don't remember getting my marks.

Siddhartha Srivastava - 6 years, 3 months ago

@Siddhartha Srivastava – There are some rumors that with the facilitation of e-correction, marksheets will be e-mailed to the parents

Krishna Ar - 6 years, 3 months ago

what do u mean by "badly flunked" u r Level 5 in Combi

Adarsh Kumar - 6 years, 3 months ago

I hate calculations. End up making silly mistakes, end up flunking exams :3

Krishna Ar - 6 years, 3 months ago

@Krishna Ar – Same here!!

Adarsh Kumar - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$