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Simple dice rolls but one has adding probabilities and anoth

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Simple dice rolls but one has adding probabilities and anoth

by manimgoindowndown » Sat Jan 19, 2013 11:46 am
er has multiplying them? Why?

The first one involves adding them second one involves multiplying the probabilities of not getting them and subtracting them by one.

molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. what is the probability that molly will roll the die less than four times before stopping?

This makes sense to me to add

But how abou this one
What is the probability that, on three rolls of a single fair die, AT LEAST ONE of the rolls will be six?

To me, I don't really see how the situation changed in the second problem. All three dice rolls are still mututally exlcusive( adding probabilities) in the second problem
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by Tommy Wallach » Sat Jan 19, 2013 3:54 pm
Hey Mani,

So here's the thing about probabilities. Don't think of it as an "addition" question or a "multiplication" question. You just need to remember that anytime you need more than one thing to happen AT THE SAME TIME, you multiply. Any time you are okay with two things happening but not at the same time, you add them.

1)

odds of getting one on the first roll + odds of getting something other than one on the first roll and then getting one on the second roll + odds of getting something other than one on the first two rolls and then getting one on the third roll.

In this example, there are THREE TOTALLY SEPARATE SITUATIONS that meet the criterion requested (stopping before four rolls)

1/6 + 5/6 * 1/6 + 5/6 * 5/6 * 1/6

Notice that we are both adding AND multiplying to solve here.

2)

1 - (probability of No sixes at all)

In this example, there is only ONE situation that meets the criterion requested, so we only need to calculate one probability (the issue of subtracting from 1 is only because this second example uses the language "at least," which the first one does not).

Keep in mind, the dice rolls in the first problem are NOT mutually exclusive, because the only way you can get to the second roll is if the first roll is NOT a one. This means those rolls are interdependent.

Does that help?

-t
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by manimgoindowndown » Sat Jan 19, 2013 9:33 pm
I think I understood it better. Conceptually I just don't understand why we're multiplying in the second problem. I know from practice, but conceptually it's still unclear why we're multiplying. Is it because to me whatever you roll on each roll is independent on each roll?

Also we calculate P(1)+p(2) and added them, because using your concept...you can't have P exactly 1 P exactly 2 occur at the same time. It's EITHER OR of those situations occuring.



This was really helpful "Keep in mind, the dice rolls in the first problem are NOT mutually exclusive, because the only way you can get to the second roll is if the first roll is NOT a one. This means those rolls are interdependent." And just spells out what I intuitively know

Thanks for the time and help man. I really appreciate it.

Part of the reason I might have gotten confused was I build the schema for independent (multiplication) and mutually exlcusive (addition) based on coins and dice

flipping two coins and getting heads you multiply 1/2 x 1/2
The events, on one flip of getting EITHER a heads or tails is mutually exlcusive from each other. I wonder what a simple problem for this would be?


rolling a dice and getting two sixes twice would be 1/6x1/6 (same concept as above). I wonder what a good dice example would be for mutually exclusive probablities?

Conceptually it wouldn't be addition because adding probabilities we'd quickly get to 1 or 100%

I am just trying to get a conceptual problem to start the fundmanetal for both of them to build off of. That or my head is fried from doing so many problems that I tend to know which one intuitively, but I don't want to go off of intuition


I know this post is a little disorganized and this is a slightly different topic, but I noticed on rate problems when the langauge is stated like this
https://www.manhattangmat.com/forums/lia ... 14082.html

Real situation- rate
Hypothetical situation- Liam goes 5mph slower

that instead of portrying the rate of a hypothetical situation as r-5, most gmat books portray the hypothetical as r and the actual as r+5. That seems a little counterintuitive to me except that it's preferrabl to have a positive value for ate
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by Tommy Wallach » Sat Jan 19, 2013 9:42 pm
In the second problem, it's 1 - 5/6 * 5/6 * 5/6

You multiply whenever you need multiple things to happen. To solve for no sixes on three rolls, you need to get something other than a six on every roll. In the first question, you have to do 5/6 * 1/6 because you need to get a non-one on the first throw AND a one on the second, etc. Any time you need two things to happen, multiply. Think of it this way; probabilities other than 100% are all less than one. The probability of ONE thing happening must be either greater than or equal to the probability of that one thing AND another thing happening. Fractions get smaller when you multiply...so multiply.

Does that help?

T

Ps. As for your rate question, it should never matter which way you choose to write it. Has there been some kind of problem when you write it the other way?
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by manimgoindowndown » Sun Jan 20, 2013 2:32 am
Hey thanks it helps. Another way I kind of conceptualized this after thinking about it is maybe to look at it like sets. A mutually exlcusive problem has no two intersecting sets. You can't flip both heads and tails on a coin, but then when I want two heads in a row I will look at a situation as independent of another, and having a seperate outcome ie HT vs TH and adding those probabilities.

Thanks again
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