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.

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TopNewest@Azhaghu Roopesh M @Satvik Golechha – Adarsh Kumar · 2 years, 7 months ago

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– Satvik Golechha · 2 years, 7 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.Log in to reply

– Adarsh Kumar · 2 years, 7 months ago

oh!thanxxx!!Log in to reply

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 . – Azhaghu Roopesh M · 2 years, 7 months ago

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– Adarsh Kumar · 2 years, 7 months ago

Hey!No problem man!Log in to reply

LOL this was a question in my model exam which I badly flunked >:o – Krishna Ar · 2 years, 7 months ago

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– Adarsh Kumar · 2 years, 7 months ago

what do u mean by "badly flunked" u r Level 5 in CombiLog in to reply

– Krishna Ar · 2 years, 7 months ago

I hate calculations. End up making silly mistakes, end up flunking exams :3Log in to reply

– Adarsh Kumar · 2 years, 7 months ago

Same here!!Log in to reply

– Azhaghu Roopesh M · 2 years, 7 months ago

BTW Which model exam , do you mean for Boards ?Log in to reply

– Krishna Ar · 2 years, 7 months ago

Yup! :PLog in to reply

– Siddhartha Srivastava · 2 years, 7 months ago

Aren't you in CBSE? They still have the CGPA system right? Don't Worry. Everyone gets CGPA 10. Everyone.Log in to reply

– Krishna Ar · 2 years, 7 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.Log in to reply

– Siddhartha Srivastava · 2 years, 7 months ago

I know, I benefited from the same. :DLog in to reply

– Krishna Ar · 2 years, 7 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)Log in to reply

I thought they only tell you your grades? Not your marks. I don't remember getting my marks. – Siddhartha Srivastava · 2 years, 7 months ago

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– Krishna Ar · 2 years, 7 months ago

There are some rumors that with the facilitation of e-correction, marksheets will be e-mailed to the parentsLog in to reply

– Azhaghu Roopesh M · 2 years, 7 months ago

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

– Krishna Ar · 2 years, 7 months ago

Alias. Not too keen on revealing my name(at least as of now)Log in to reply

– Azhaghu Roopesh M · 2 years, 7 months ago

Fine with me ,I was just curious :)Log in to reply

– Krishna Ar · 2 years, 7 months ago

haha :)Log in to reply

If you want , you can visit this link to see the names :) – Azhaghu Roopesh M · 2 years, 7 months ago

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– Siddhartha Srivastava · 2 years, 7 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.Log in to reply

– Azhaghu Roopesh M · 2 years, 7 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:)Log in to reply

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 · 2 years, 7 months ago

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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 · 2 years, 7 months ago

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– Roman Frago · 2 years, 7 months ago

The analogy does not apply...Log in to reply

If the two dice can be distinguished from one another, they become independent events. – Roman Frago · 2 years, 7 months ago

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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 · 2 years, 7 months ago

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– Roman Frago · 2 years, 7 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.Log in to reply

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.

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 · 2 years, 7 months ago

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Addendum: With distinguishable dice, you can identify 2-1 from 1-2, even 1-1 from 1-1 making them additional outcome. – Roman Frago · 2 years, 7 months ago

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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 · 2 years, 7 months ago

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– Roman Frago · 2 years, 7 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?Log in to reply

– Azhaghu Roopesh M · 2 years, 7 months ago

What do you mean by "OP" ?Log in to reply

– Siddhartha Srivastava · 2 years, 7 months ago

Original poster.Log in to reply

– Adarsh Kumar · 2 years, 7 months ago

Thanxx!I got it!Log in to reply