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# Math: Probability

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Re: Math: Probability [#permalink]

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10 Jul 2013, 23:07
Bumping for review*.

*New project from GMAT Club!!! Check HERE

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Re: Math: Probability [#permalink]

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26 Dec 2013, 05:09
fruit wrote:
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?)

Are there any GMAT questions testing this concept?

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Re: Math: Probability [#permalink]

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26 Dec 2013, 05:29
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Re: Math: Probability [#permalink]

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27 Dec 2013, 00:36
Thanks Bunuel

Thanks for the prompt reply

+1

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Re: Math: Probability [#permalink]

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12 Mar 2014, 10:22
Dear Walker

I was trying your given example- 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? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help!

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Re: Math: Probability [#permalink]

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13 Mar 2014, 01:00
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ShantnuMathuria wrote:
Dear Walker

I was trying your given example- 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? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help!

The following post might help: if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html#p764040

Also check similar questions to practice:
a-committee-of-three-people-is-to-be-chosen-from-four-teams-130617.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
a-comittee-of-three-people-is-to-be-chosen-from-four-married-130475.html
a-committee-of-three-people-is-to-be-chosen-from-4-married-101784.html
a-group-of-10-people-consists-of-3-married-couples-and-113785.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
given-that-there-are-6-married-couples-if-we-select-only-58640.html
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Re: Math: Probability [#permalink]

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03 Dec 2014, 23:14
walker

Can you help me solve below example by reversal probability approach?

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?

Thanks!
Anupam

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Re: Math: Probability [#permalink]

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04 Dec 2014, 03:31
AnuLalwani wrote:
walker

Can you help me solve below example by reversal probability approach?

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?

Thanks!
Anupam

This question is discussed here: if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html

Hope it helps.
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23 May 2015, 06:49
frank1 wrote:
walker 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.

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

can somebody explain this to me...
seems bit complicated....

Total number of ways to select 3 people from 5 couples = 10C3

Number of ways to select 3 people such that there is no married couple = (Total number of ways) - (Ways to form a committee in which a married couple is there)
= 120 - (5C1 x 8C1)

5C1 -> select 1 married couple
8C1 -> select any guy other than the married couple selected

Number of ways to select 3 people such that there is no married couple = 120 - 40 = 80

Probability of choosing 3 people so that there is no married couple = 80/120 = 2/3

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Re: Math: Probability [#permalink]

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Re: Math: Probability [#permalink]

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01 Aug 2016, 22:40
a box contains 100 balls, numbered from 1 to100.if 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on balls selected from the box will be odd

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Re: Math: Probability [#permalink]

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01 Aug 2016, 23:10
beney81 wrote:
a box contains 100 balls, numbered from 1 to100.if 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on balls selected from the box will be odd

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This is not the way a question should be posted on the forum. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Thank you.

This question is discussed here: http://gmatclub.com/forum/a-box-contain ... 09279.html
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Re: Math: Probability [#permalink]

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22 Oct 2016, 03:43
abagnale11 wrote:
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

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

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

Hi Thank you very much for these information. I found it very interesting. However, I need your help on your conclusion: X+Y+Z=10, should the solution be 12C2 instead of 14C4? I'm confused.

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Re: Math: Probability [#permalink]

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25 Feb 2017, 21:59
"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′=pk∗(1−p)n−kP′=pk∗(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=Cnk∗pk∗(1−p)n−kP=Ckn∗pk∗(1−p)n−k (2)

In the example with a coin, right answer is P=C83∗0.53∗0.55=C83∗0.58P=C38∗0.53∗0.55=C38∗0.58

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=C72∗0.42∗0.65P=C27∗0.42∗0.65"

I didn't follow the part of the formula where the probability of it not raining is taken into account. The formula counts all combinations where all the desired outcomes and its probability of occurring (0.4^2). I get that the opposite possibility must also be taken into account but I'm not able to follow the formula. Can someone explain?

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Re: Math: Probability [#permalink]

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25 Feb 2017, 22:05
"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′=pk∗(1−p)n−kP′=pk∗(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=Cnk∗pk∗(1−p)n−kP=Ckn∗pk∗(1−p)n−k (2)

In the example with a coin, right answer is P=C83∗0.53∗0.55=C83∗0.58P=C38∗0.53∗0.55=C38∗0.58

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 #a2:
P=C72∗0.42∗0.65P=C27∗0.42∗0.65"

I didn't follow the part of the formula where the probability of it not raining is taken into account. The formula counts all combinations where all the desired outcomes and its probability of occurring (0.4^2). I get that the opposite possibility must also be taken into account but I'm not able to follow the formula. Can someone explain?

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Re: Math: Probability [#permalink]

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25 Feb 2017, 22:21
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).

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

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

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?

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Re: Math: Probability [#permalink]

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26 Feb 2017, 04:51
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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?

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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05 Jul 2017, 14:26
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?

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Re: Math: Probability [#permalink]

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16 Jul 2017, 04:36
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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 [ 136.95 KiB | Viewed 618 times ]

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Re: Math: Probability [#permalink]

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16 Jul 2017, 18:26
ganand

Thank you, that helps. Got it.

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Re: Math: Probability   [#permalink] 16 Jul 2017, 18:26

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# Math: Probability

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