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Probability Basics Rules & Questions (43)

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Probability Basics Rules & Questions (43) [#permalink]

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Normally I have heard that Probability Questions are tough so here are few basic Combinatory and probability questions 43 of them so that you can have some practice for them.

Practice makes a man perfect.............!!!!!! :read

Don't forget to give a Kudos if this Document helps ...:thumbup:


Probability-Practice - _Good_ones.pdf owner is Qweert

Don't forget to give him a Kudos if this Document helps ...:thumbup:
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Re: Probability Basics Rules & Questions (43) [#permalink]

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Probability
Overview

In this section we will cover:

Expressing probabilities
GMAT Probability Rule #1
GMAT Probability Rule #2
GMAT Probability Rule #3
GMAT Probability Rule #4

Probability
You may encounter questions on the GMAT that ask the probability of an event occuring. For instance:
• Sara rolls two fair, six-sided dice. What is the probability she will roll two 6s?
• Roland flip two fair coins. What is the probability that neither of the coins will land on tails?
• A bag contains six blue marbles and six red marbles. If Teresa randomly chooses a marble from the bag, what is the probability that the marble is blue?
All of these problems are probability problems. Probability is a measure of the likelihood of an event occurring.
Expressing Probability
The probability of an event occurring is expressed as a number between 0 and 1. A probability of 0 means the event can never happen. A probability of 1 means the event is certain to happen.

You may see probability express on the GMAT in terms of a percent, a decimal, or a fraction. For instance, the probability of a fair coin landing on tails can be expressed as


Fraction 1/2
Decimal .5
Percent 50%

Sometimes you may see notation similar to that below:

P(A) = 1/2

This means that the probability of an event A is 1/2.

Probability Rules
There are four probability rules you need to memorize in order to master GMAT probability questions:

1. The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

2. The probability of an event occurring plus the probability of the event not occurring equals 1.

3. The probability of event A AND event B occurring is the probability of event A times the probability of event B given that event A has already occurred.

4. The probability of event A OR event B occurring is the probability of event A occurring plus the probability of event B occurring minus the probability of both events occurring.
The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

Example: Raphael tosses a fair coin. What is the probability the coin will come up heads?

Probability of heads = [heads]/[heads, tails]
Probability of heads = 1/2

Example: Tom rolls a fair die. What is the probability that the die will roll an even number?

Probability of even number = [2, 4, 6]/[1, 2, 3, 4, 5, 6]
Probability of even number = 3/6
Probability of even number = ½

Rule #2
The probability of an event occurring plus the probability of the event not occurring equals 1.

In other words, we can say with 100% certainty that an event will either occur or not occur. For instance, the probability of a fair, six-sided die rolling a 4 is 1/6. The probability of the die not landing on 4 is (1 - 1/6) or 5/6. 1/6 + 5/6 = 1.

This concept can be very helpful on the GMAT. Sometimes it is easier to determine the probability of an event not occurring than determining the probability of an event occurring. Once your know the probability of an event not occurring, you can subtract the probability from 1 to find the probability of an event occurring.

Rule #3
The probability of event A AND event B occurring is the probability of event A times the probability of event B given that event A has already occurred.

Example: Joseph rolls two fair, six-sided die. What is the probability that both die will roll a 6?

Probability of 1st die coming up 6: 1/6
Probability of 2nd die coming up 6: 1/6
Probability of both die coming up 6: (1/6) * (1/6)
Probability of both die coming up 6: 1/36

Example: A bag contains three blue marbles and three red marbles. If two marbles are drawn randomly from the bag, what is the probability that they are both blue?

This problem is a dependent probability. Two events are said to be dependent events if the outcome of one event affects the outcome of the other event. The probability of drawing the second marble depends on the outcome of the first marble. If the first marble is red, there is no possibility of drawing two blue marbles. Thus, the probability of drawing a second blue marble is calculated after the first blue marble has been drawn.

Probability of drawing blue on first draw: 3/6

If a blue is drawn on the first draw, there are three red marbles and two blue marbles remaining in the bag.

Probability of drawing blue on second draw (given that first was blue): 2/5

Probability of drawing two blue: (3/6) * (2/5)
Probability of drawing two blue: 6/30
Probability of drawing two blue: 1/5

Rule #4
The probability of event A OR event B occurring is the probability of event A occurring plus the probability of event B occurring minus the probability of both events occurring.

Example: Charles rolls a fair, six-sided die. What is the probability of Charles rolling a 2 or a 4?

Probability of 2: 1/6
Probability of 4: 1/6
Probability of a 2 or 4: 1/6 + 1/6
Probability of a 2 or 4: 2/6
Probability of a 2 or 4: 1/3

In the previous problem, the events were mutually exclusive. Mutually exclusive means that the events cannot occur together. There is no way to roll a 2 and a 4 at the same time. The events in the following problem are NOT mutually exclusive.

Example: Of the 100 students at a certain school, 30 students are taking a chemistry class, 40 students are taking a physics class, and 20 students are taking both a physics and a chemistry class. If a student is chosen at random from the school, what is the probability that he or she is taking a physics or a chemistry class?

Probability of selecting a student taking a chemistry course: 30/100
Probability of selecting a student taking a physics class: 40/100
Probability of selecting a student taking both classes: 20/100
Probability of a selecting a student taking chemistry OR a student taking physics: 30/100 + 40/100 - 20/100 = 50/100 or 1/2
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Re: Probability Basics Rules & Questions (43) [#permalink]

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Here are some good probability problems...
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Re: Probability Basics Rules & Questions (43) [#permalink]

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New post 22 Aug 2010, 05:48
Let us make this post Nightmare for Probability Questions and heavens for people afraid of it.....

I will work hard you will work hard but Together we will definitely Succeed
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I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
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Re: Probability Basics Rules & Questions (43) [#permalink]

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Hey Qweert I am adding this doc to my initial post so we have everything in the main post
I hope you don't have any issue...!!!
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Re: Probability Basics Rules & Questions (43) [#permalink]

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New post 30 Oct 2010, 00:25
thanks a lot for this! probability for me was really confusing and your notes were a great help for my PE Exam review! ;)

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Re: Probability Basics Rules & Questions (43) [#permalink]

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Re: Probability Basics Rules & Questions (43) [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: Probability Basics Rules & Questions (43) [#permalink]

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New post 24 Sep 2017, 16:36
good questions, but I'll be careful with the provided answers and the questions wording. number 4 for example. How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?
in GMAT when you say and the other digits ( it means the 1st digit is not included). Similarly, are all different from each other came after "the other digits" so only the two remaining digits are different from each others. note that the word all here is wrong as "all" should indicates more than 2.
Also the 1st digit means the 1st digit from the right and not vice versa.

if I was to answer this question, my answer will be as following:
1st digit can only takes 9. no condition
2nd &3rd:
2nd digit can take 10 (0-9)
3rd can take only 9 (1-9) since zero won't make the number 3 digits.
from the 2nd &3rd digits you need to subtract the similar numbers which is 9 (1-9) since zero can only be assigned to the 2nd digit.

so now you have 1st(9 options) * 2nd&3rd(9x10-9)=81
answer is 9*81=729

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Re: Probability Basics Rules & Questions (43) [#permalink]

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New post 24 Sep 2017, 19:11
Number.37

there is a scenario not taken into account, 4th and 5th digit from the left equal each others. this should be added to the answer

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Re: Probability Basics Rules & Questions (43) [#permalink]

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New post 26 Sep 2017, 08:35
This question set has the answers typed out wrong. You must read the explanation to get the right answer choice rather than simply reading the letter mentioned in the document...

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Re: Probability Basics Rules & Questions (43)   [#permalink] 26 Sep 2017, 08:35
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