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# 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.

Note by Adarsh Kumar
2 years, 5 months ago

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@Azhaghu Roopesh M @Satvik Golechha · 2 years, 5 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. · 2 years, 5 months ago

oh!thanxxx!! · 2 years, 5 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 . · 2 years, 5 months ago

Hey!No problem man! · 2 years, 5 months ago

LOL this was a question in my model exam which I badly flunked >:o · 2 years, 5 months ago

what do u mean by "badly flunked" u r Level 5 in Combi · 2 years, 5 months ago

I hate calculations. End up making silly mistakes, end up flunking exams :3 · 2 years, 5 months ago

Same here!! · 2 years, 5 months ago

BTW Which model exam , do you mean for Boards ? · 2 years, 5 months ago

Yup! :P · 2 years, 5 months ago

Aren't you in CBSE? They still have the CGPA system right? Don't Worry. Everyone gets CGPA 10. Everyone. · 2 years, 5 months ago

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. · 2 years, 4 months ago

I know, I benefited from the same. :D · 2 years, 4 months ago

Haha , now my parents know about this, so they want a cgpa 10 w/o upgrades. (in fact, they want centums too :3) · 2 years, 4 months ago

(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. · 2 years, 4 months ago

There are some rumors that with the facilitation of e-correction, marksheets will be e-mailed to the parents · 2 years, 4 months ago

Hey Krishna , I wanted to ask this for some time now , is Krishna Ar your real name or is it an alias ? · 2 years, 4 months ago

Alias. Not too keen on revealing my name(at least as of now) · 2 years, 4 months ago

Fine with me ,I was just curious :) · 2 years, 4 months ago

haha :) · 2 years, 4 months ago

Agreed !! 2 years ago there were 71 students who got 10 CGPA !!

If you want , you can visit this link to see the names :) · 2 years, 5 months ago

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. · 2 years, 4 months ago

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:) · 2 years, 4 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. · 2 years, 5 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 $$\frac{1}{2}$$ · 2 years, 5 months ago

The analogy does not apply... · 2 years, 4 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. · 2 years, 4 months ago

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. · 2 years, 4 months ago

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. · 2 years, 4 months ago

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. · 2 years, 4 months ago

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. · 2 years, 4 months ago

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. · 2 years, 4 months ago

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? · 2 years, 4 months ago

What do you mean by "OP" ? · 2 years, 5 months ago

Original poster. · 2 years, 4 months ago