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Math: Probability

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Math: Probability [#permalink]

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PROBABILITY

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This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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Definition

A number expressing the probability (p) that a specific event will occur, expressed as the ratio of the number of actual occurrences (n) to the number of possible occurrences (N).

\(p = \frac{n}{N}\)

A number expressing the probability (q) that a specific event will not occur:

\(q = \frac{(N-n)}{N} = 1 - p\)

Examples

Coin

ImageImage

There are two equally possible outcomes when we toss a coin: a head (H) or tail (T). Therefore, the probability of getting head is 50% or \(\frac{1}{2}\) and the probability of getting tail is 50% or \(\frac{1}{2}\).
All possibilities: {H,T}

Dice

Image

There are 6 equally possible outcomes when we roll a die. The probability of getting any number out of 1-6 is \(\frac{1}{6}\).
All possibilities: {1,2,3,4,5,6}

Marbles, Balls, Cards...

Image

Let's assume we have a jar with 10 green and 90 white marbles. If we randomly choose a marble, what is the probability of getting a green marble?
The number of all marbles: N = 10 + 90 =100
The number of green marbles: n = 10
Probability of getting a green marble: \(p = \frac{n}{N} = \frac{10}{100} = \frac{1}{10}\)

There is one important concept in problems with marbles/cards/balls. When the first marble is removed from a jar and not replaced, the probability for the second marble differs (\(\frac{9}{99}\) vs. \(\frac{10}{100}\)). Whereas in case of a coin or dice the probabilities are always the same (\(\frac{1}{6}\) and \(\frac{1}{2}\)). Usually, a problem explicitly states: it is a problem with replacement or without replacement.


Independent events

Two events are independent if occurrence of one event does not influence occurrence of other events. For n independent events the probability is the product of all probabilities of independent events:

p = p1 * p2 * ... * pn-1 * pn

or

P(A and B) = P(A) * P(B) - A and B denote independent events

Example #1
Q:There is a coin and a die. After one flip and one toss, what is the probability of getting heads and a "4"?
Solution: Tossing a coin and rolling a die are independent events. The probability of getting heads is \(\frac{1}{2}\) and probability of getting a "4" is \(\frac{1}{6}\). Therefore, the probability of getting heads and a "4" is:
\(P = \frac{1}{2} * \frac{1}{6} = \frac{1}{12}\)

Example #2
Q: If there is a 20% chance of rain, what is the probability that it will rain on the first day but not on the second?
Solution: The probability of rain is 0.2; therefore probability of sunshine is q = 1 - 0.2 = 0.8. This yields that the probability of rain on the first day and sunshine on the second day is:
P = 0.2 * 0.8 = 0.16

Example #3
Q:There are two sets of integers: {1,3,6,7,8} and {3,5,2}. If Robert chooses randomly one integer from the first set and one integer from the second set, what is the probability of getting two odd integers?
Solution: There is a total of 5 integers in the first set and 3 of them are odd: {1, 3, 7}. Therefore, the probability of getting odd integer out of first set is \(\frac{3}{5}\). There are 3 integers in the second set and 2 of them are odd: {3, 5}. Therefore, the probability of getting an odd integer out of second set is \(\frac{2}{3}\). Finally, the probability of of getting two odd integers is:
\(P = \frac{3}{5} * \frac{2}{3} = \frac{2}{5}\)


Mutually exclusive events

Shakespeare's phrase "To be, or not to be: that is the question" is an example of two mutually exclusive events.

Two events are mutually exclusive if they cannot occur at the same time. For n mutually exclusive events the probability is the sum of all probabilities of events:

p = p1 + p2 + ... + pn-1 + pn

or

P(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events

Example #1
Q: If Jessica rolls a die, what is the probability of getting at least a "3"?
Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is:
\(P = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3}\)


Combination of independent and mutually exclusive events

Many probability problems contain combination of both independent and mutually exclusive events. To solve those problems it is important to identify all events and their types. One of the typical problems can be presented in a following general form:

Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)
Solution: All events are independent. So, we can say that:

\(P' = p^k*(1-p)^{n-k}\) (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails:
HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\) (2)

In the example with a coin, right answer is \(P = C^8_3*0.5^3*0.5^5 =C^8_3*0.5^8\)

Example #1
Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period?
Solution: We are not interested in the exact sequence of event and thus apply formula #2:
\(P = C^7_2*0.4^2*0.6^5\)


A few ways to approach a probability problem

There are a few typical ways that you can use for solving probability questions. Let's consider example, how it is possible to apply different approaches:

Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

1) combinatorial approach: The total number of possible committees is \(N=C^8_2\). The number of possible committee that includes both Bob and Rachel is \(n=1\).
\(P = \frac{n}{N} = \frac{1}{C^8_2} = \frac{1}{28}\)

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is \(N=C^8_2\). The number of possible committee that does not includes both Bob and Rachel is:
\(m = C^6_2 + 2*C^6_1\) where,
\(C^6_2\) - the number of committees formed from 6 other people.
\(2*C^6_1\) - the number of committees formed from Rob or Rachel and one out of 6 other people.
\(P = 1- \frac{m}{N} = 1 - \frac{C^6_2 + 2*C^6_1}{C^8_2}\)
\(P = 1 - \frac{15+2*6}{28} = 1 - \frac{27}{28} = \frac{1}{28}\)

3) probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is.
\(P = \frac{2}{8} * \frac{1}{7} = \frac{2}{56} = \frac{1}{28}\)

4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * 1) = \frac{2}{56} = \frac{1}{28}\)

Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


Probability tree

Sometimes, at 700+ level you may see complex probability problems that include conditions or restrictions. For such problems it could be helpful to draw a probability tree that include all possible outcomes and their probabilities.

Example #1
Q: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?
Solution: Let's draw all possible outcomes and calculate all probabilities.

Image

Now, It is pretty obvious that the probability of Julia's win is:
\(P = \frac47 + \frac27*\frac16 + \frac17*\frac26 = \frac23\)


Tips and Tricks: Symmetry

Symmetry sometimes lets you solve seemingly complex probability problem in a few seconds. Let's consider an example:

Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?
Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly 1/2 of all possible ways:
\(N = \frac12*P^5_2 = 10\)


Official GMAC Books:

The Official Guide, 12th Edition: DT #4; DT #7; PS #12; PS #67; PS #105; PS #158; PS #174; PS #214; DS #3; DS #107;
The Official Guide, Quantitative 2th Edition: PS #79; PS #160;
The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;

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Resources

Probability DS problems: [search]
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Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]
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Last edited by walker on 25 Jan 2015, 09:39, edited 25 times in total.
Math expressions & minor clarifications

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Man....it must have taken lots of efforts to compile this document.....

kudos to u...


Thanx for sharing this...
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jasonedward wrote:
I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14
Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.


Donald to receiver either Prograine or Ropecia must be among first two chosen patients and as there are 14 patients then the probability of this is simply 2/14=1/7.
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You are right! I need to double check each example!
+1

Good luck on exam!!!
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hgp2k wrote:
walker wrote:
......
P = 1/6 + 1/6 + 1/6 = 1/2 - I think this should be 2/3, as there are 4 number which satisfies the condition (at least 3) {3, 4, 5, 6}.


Another excellent post by Walker!!! +1 :)


Thanks! You are absolutely right.
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Wow - between Walker and Bunuel, we have Math covered.
Great job!
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Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?

Slot method to solve this problem:

Chair location: 1 2 3 4 5


R has 4 possible choices (2, 3, 4, 5).

When R is on chair 5, B can have 4 options to choose from (1, 2, 3, 4) that are left of chair 5 = 4
When R is on chair 4, B can have 3 options to choose from (1, 2, 3) that are left of chair 4 = 3
When R is on chair 3, B can have 2 options to choose from (1, 2) that are left of chair 3 = 2
When R is on chair 2, B can have 1 option to choose from (1) that is left of chair 2 = 1

Possible ways = 4 + 3 + 2 + 1 = 10

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walker wrote:
Mutually exclusive events

Two events are mutually exclusive if they cannot occur at the same time. For n mutually exclusive events the probability is the sum of all probabilities of events:

p = p1 + p2 + ... + pn-1 + pn

or

P(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events

Example #1
Q.:Jessica rolls a dice. What is probability of getting at least 3?
Solution: There are 3 outcomes that satisfy our condition (at least 3): {1, 2, 3}. The probability of each outcome is 1/6. The probability of getting at least 3 is:
P = 1/6 + 1/6 + 1/6 = 1/2 - I think this should be 2/3, as there are 4 number which satisfies the condition (at least 3) {3, 4, 5, 6}.


Another excellent post by Walker!!! +1 :)

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Thanks, great post!

One question on the last example. Isn't 5P2 equal to 20? n!/(n-r)! would be 5!/3!=5*4=20.
So 1/2 of 5P2 would actually be 10 not 30.

more specifically, the 10 ways Bob and Rachel could sit would be

B R _ _ _
B_ R _ _
B _ _ R _
B _ _ _ R
_ B R _ _
_ B _ R _
_ B _ _ R
_ _ B R _
_ _ B _ R
_ _ _ B R

I have my test in two days, and not really sure why I am still studying :?

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Thanks for both, theory and examples.

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\(C^6_1\) - the number of committees formed from Bob and one person out of 6 people.
\(C^6_1\) - the number of committees formed from Rachel and one person out of 6 people.

So, \(C^6_1 + C^6_1 = 2*C^6_1\) - the number of committees formed from Rob or Rachel and one person out of 6 other people.
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jasonedward wrote:
I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13-- something wrong here

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14
Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.


Donald Getting prograine = 1/14 (1 out of 14 patients should be chosen for prograine)
Donald Not getting prograine = 13/14
Donald Getting Ropecia = 1/13 (1 out of 13 patients should be chosen for Ropecia)

How can Donald get either P or R
Donald will get Prograine and get done OR Donald will not get Prograine and get Ropecia

1/14+ 13/14*1/13 = 1/14+1/14 = 1/7

I believe it is a without replacement problem.
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Let me briefly talk about a concept that i remember studying long back.Most of the logic would be excerpts from Higher Algebra by Bernard Shaw

We know that most of probability questions can be easily cracked if we have a strong grip on Permutation and combination.Let me present you all with something interesting. I'll be dealing with "Distribution" of similar and dissimilar things

Concept 1: Distributing Similar things among people where each recipient receives at least 1 thing
----------------------------------------------------------------------------------------------

Imagine you have 10 similar tennis balls which you need distribute it to 3 friends and each of the fella receives at least 1 ball.

how would you approach it???

Let us do it figuratively.Imagine the 10 balls

O O O O O O O O O O ( remember there are 9 spaces between 10 balls)

if you consider 2 sticks ! !, how many ways can u place them in the 9 spaces between the 10 balls. yeah right 9C2 ways.. would u believe , thats the answer. for example consider O O ! O O O ! O O O O O . this is one of the 9C2 ways where the first guy gets 2 balls, 2 nd guy gets 3 and the last gets 5. you just need to place (r-1) sticks in the spaces between the n balls and that's like selecting r-1 spaces between the n balls.
so the answer is n-1Cr-1 n= number of balls r= number of recepients
I can rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 9C 2 :shock:


Concept 2 : Distributing Similar things among people where each recipient may receive no balls at all
----------------------------------------------------------------------------------------------

Imagine the same old story where you have 10 similar tennis balls which you need distribute it to 5 friends. However now 1 or more of them may receive 0 balls

Again imagine the balls lying beautifully in the imagination of your brain

O O O O O O O O O O .

Now as you know for 5 friends you need to consider 4 sticks ! ! ! !. Now that any of the friend may receive 0 balls also, we cannot consider the space between the balls. So what do you think.. let me figuratively sample u some solution

one of the solution O ! O ! O O ! O O O ! O O O ( FRIEND 1= 1 BALL, FRIEND 2= 1 BALL FRIEND 3= 2 BALLS FRIEND 4= 3 BALLS FRIEND 5 = 3BALLS)
ANOTHER SOLUTION ! O O O ! O O O ! O O O O ! ( FRIEND 1 = 0 BALLS , FRIEND 2= 3 BALLS , FRIEND 3= 3 BALLS , FRIEND 4 = 4 BALLS FRIEND 5 = 0)

I hope by now you would have guessed the solution is nothing but arranging 14 things of which 10 are of one kind and 4 are of one kind

that is !14/(!10*!4) . this is nothing but n+r-1 C n where n is the number of balls and r is the number of recipients

I can again rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >=0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 14C4 :shock:



I can present some more topics. this will be based on the condition that the readers find it interesting and helpful

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Re: Math: Probability [#permalink]

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Aditi10 wrote:
Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

--------------------------------------------------------


Didn't follow the reverse combinatorial approach here. The question asks the probability that none of the 3 people selected would be married. But the approach is taking 1 married couple and 1 unmarried person. What am I missing?


Aditi10,

We can select 3 people out of 10 peoples (5 married couples) in two possible ways:

1. Event A: None of them are married to each other
or
2. Event B: 1 married couple and 1 person.

Note that event A and B are mutually exclusive and exhaustive events.

We know that total probability is always 1. => P(A) + P(B) = 1

In the first approach, we directly compute P(A).

In "reverse combinatorial" approach, we calculate first P(B) and then subtract it from 1 to get P(A).

Hope it helps.

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Re: Math: Probability [#permalink]

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Laxmib wrote:
Can any one help me solve below example and how the probability tree works?

Q: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?


Hi Laxmib,

Please find attached probability tree for the above problem.
Required answer = 4/7 + 2/42 + 2/42 = 28/42 = 2/3 .

Hope this helps.

Thanks.
Attachments

Probability_Tree_Julia_Brian.jpg
Probability_Tree_Julia_Brian.jpg [ 136.95 KiB | Viewed 423 times ]

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Re: Math: Probability [#permalink]

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azelastine wrote:
This is very basic but I am getting confused by the concepts.

If a fair coin is flipped three times, what is the probability that it
comes up heads all three times?

So the way I understand it, P of getting tails is 1/2*1/2*1/2=1/8

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

What am I doing wrong?


Hi azelastine,

Your first part is correct, i.e. probability of getting 3 tails is 1/8.
Similarly, you can compute the probability of getting three heads.
Please note that each coin toss is independent of one another, that's why we can multiply the probability of each event.
P(HHH) = P(getting head on the first toss) and P(getting head on the second toss) and P(getting head on the third toss)= 1/2 * 1/2 * 1/2 = 1/8

AND implies Multiplication(*)
OR implies Addition(+)

Quote:
Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....


Here you are assuming that following events are complementary to each other:
1. Getting three tails, i.e. P(TTT) and
2. Getting three heads, i.e.P(HHH)
P(TTT) + P(HHH) = 1
This is not correct.

Let's enumerate the all possible outcomes(sample space).
1. HHH
2. HHT
3. HTT
4. TTT
5. TTH
6. THH
7. THT
8. HTH

Sum of probabilities of all these events will be equal to 1. i.e.
P(HHH) + P(HHT) + P(HTT) + P(TTT) + P(TTH) + P(THH) + P(THT) + P(HTH) = 1

In general, if you toss the coin n times, then total possible outcomes = 2^n. In this cane, n= 3, so total outcomes = 2^3 = 8.

Hope it helps.

Thanks.

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Re: Math: Probability [#permalink]

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Quote:
Aren't we looking for the permutations here instead of the combination? aren't HHHTTTTT and HHTTTTTH the same combination but two different permutations?


Hi Aminaelm ,

Yes, you are right. Let's analyze it more closely.

No. of arrangements of HHHTTTTT = \(\frac{8!}{5!\times 3!} = {{8}\choose{3}}\) .

Hence, we are taking into account all the possible arrangements in the given formula. Hope this helps.

Thanks.

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Re: Math: Probability [#permalink]

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New post 03 Dec 2009, 16:58
rlstern00 wrote:
Thanks, great post!

One question on the last example. Isn't 5P2 equal to 20? n!/(n-r)! would be 5!/3!=5*4=20.
So 1/2 of 5P2 would actually be 10 not 30.

more specifically, the 10 ways Bob and Rachel could sit would be

B R _ _ _
B_ R _ _
B _ _ R _
B _ _ _ R
_ B R _ _
_ B _ R _
_ B _ _ R
_ _ B R _
_ _ B _ R
_ _ _ B R

I have my test in two days, and not really sure why I am still studying :?


Thanks. I was trying to figure that one out too!

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Re: Math: Probability [#permalink]

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New post 15 Feb 2010, 18:57
this is fantastic help!

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Re: Math: Probability [#permalink]

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New post 15 Feb 2010, 18:57
this is fantastic help!

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Re: Math: Probability   [#permalink] 15 Feb 2010, 18:57

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