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Math: Probability 20 Oct 2021, 13:17
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Wow - between Walker and Bunuel, we have Math covered.
Great job!

Hi folks,

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=12∗P52=10


I think I am missing something silly here. So, let's say I consider Bob & Rachel as 1 unit - with proper arrangement (meaning, BR (not RB)).
I have 4 items. No of possible arrangements is 4! = 24.

What's wrong?
Show more

The point is that "left of Rachel" does not mean immediately to the left, it means that Bob should be anywhere to the left.


Similar questions to practice:
https://gmatclub.com/forum/7-people-a-b- ... 80985.html
https://gmatclub.com/forum/mother-mary-c ... 86407.html
https://gmatclub.com/forum/six-mobsters- ... 26151.html
https://gmatclub.com/forum/in-how-many-d ... 91460.html
https://gmatclub.com/forum/goldenrod-and ... 82214.html
https://gmatclub.com/forum/meg-and-bob-a ... 58095.html
https://gmatclub.com/forum/7-people-a-b ... 80985.html

Hope it helps.
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Math: Probability 16 Aug 2022, 00:39
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Blair15
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?

Can someone pls explain how to solve this question? I didn't get the probability tree approach.
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Check here: https://gmatclub.com/forum/julia-and-br ... 98590.html Hope it helps.
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Math: Probability 18 Mar 2010, 23:06
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fruit
I didn't get how we have found 0.5. The power 3 - is it mean the number of needed H and the power 5 represent T? So, what is 0.5?
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0.5 is probability of getting head (or tail).

fruit

I want to specify, the powers 2 and 5 are the numbers of days?? I assume that 0,4 it's the likelihood of the raining and 0.6 of the sunny weather. Right?
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That's right.
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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. Why is it multiplied by 2? Can someone pls explain it to me?
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}
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walker
PROBABILITY

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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

This topic is included in GMAT ToolKit App (iPhone/iPod Touch)
[read more] [AppStore]



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

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:

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


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;

Generated from [GMAT ToolKit]


Resources

Probability DS problems: [search]
Probability PS problems: [search]

Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]
Bullet's post with probability problems: [Combined Probability Questions]
-------------------------------------------------------

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Can someone explain the reversal probability approach under the first bold face part?

Also please please explain the red text under the married couple example, how did you know to choose 3 couple out of 5? Is it because the question asks for 3 ppl not married? Im completely lost on this example and particularly on the reversal combinatorial approach......
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fruit
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?)
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Are there any GMAT questions testing this concept?
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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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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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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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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?
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azelastine

Thanks for the explanation, this is very helpful.

Could you help me understand the difference between this and the following problem?

A fair coin is flipped twice. What is the probability that the coin lands showing heads on the first flip, second flip or both?

The solution to this question is two-step: 1) P (two tails)=1/4, 2) 1-1/4=3/4
Show more

Hi azelastine,

I'll try to explain the difference.

Quote:
A fair coin is flipped twice. What is the probability that the coin lands showing heads on the first flip, second flip or both?
Show more

Sample space: {HH, HT, TH, TT}

Favorable event = {HH, HT, TH} and complementary event = {TT}

In two ways you can solve this question.

1. Compute the probability of the favorable event
In this case required probability = P(HH) + P(HT) + P(TH) =\(\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4}\)
or
2. Compute the probability of complementary event and subtract it from 1. You have solved it using this method.

In general, when a complementary event is small then use the 2nd approach.
The complementary approach can also be used in P&C questions.

Please go through the first post of the thread. These points are discussed in detail with examples.

Hope this helps.
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walker



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.

Show more

Could you please confirm if 2 in 2*C^6_1 above is the result of C^2_1?
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Bunuel
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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: https://gmatclub.com/forum/if-4-people-a ... ml#p764040

Also check similar questions to practice:
https://gmatclub.com/forum/a-committee-o ... 30617.html
https://gmatclub.com/forum/if-4-people-a ... 99055.html
https://gmatclub.com/forum/a-committee-o ... 94068.html
https://gmatclub.com/forum/if-a-committe ... 88772.html
https://gmatclub.com/forum/a-comittee-of ... 30475.html
https://gmatclub.com/forum/a-committee-o ... 01784.html
https://gmatclub.com/forum/a-group-of-10 ... 13785.html
https://gmatclub.com/forum/if-there-are- ... 99992.html
https://gmatclub.com/forum/given-that-th ... 58640.html
Show more


Total outcomes= 10C3=120

Favourable outcomes= (5C2x3C1 + 5C2x3C1 + 5C3 +5C3) = (30 + 30 + 10 + 10) = 80
[Since, if we select 2 females from 5 couple pairs then we have to choose 1 male from remaining 3 pairs=>none are married to each other ; In the same way we can choose 2 males and 1 female ; we can also choose 3 males out of 5 pairs=>none are married to each other ; again we can choose 3 females out of 5 pairs ; all the events are mutually exclusive so we add them;]

P(selected 3 ppl are not married to each other)=80/120=2/3

Bunuel Is my way of approaching this problem right??
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I am having trouble understanding the combinatorial and reversal combinatorial approach in terms of the logic of calculating the notation from this explanation.

Example #1 under "A few ways to approach a probability problem": \(P=\frac{n}{N}\) where N = \(C |_2^8\) I am not understanding at all how you go from \(C |_2^8\) to 28. Would anyone be willing to help explain what I am missing? At first I thought it was factorials \(\frac{8!}{2!}\) but that does not equal 28. I understand the regular probability approach and reversal probability approach, but right now am completely lost on the combinatorial.

Any help explaining would be greatly appreciated!
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walker
PROBABILITY
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:


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?
Show more

I have a doubt in example 1 and 2, both related reversal probability approach. Can you please point the mistake in my method below:
Example 1
Case 1: One person out of Rachel or Bob, then other person can be anyone
Case 2: Any person in group except Rachel or Bob.

Prob. of case 1: \(\frac{2}{8} * \frac{6}{7} \)- one person either Bob or Rachel - 2 out of 8, second person anyone except Bob or Rachel - 6 out of 7
Prob. of case 2: \(\frac{6}{8} * \frac{5}{7} \)- one person anyone except Bob or Rachel - 6 out of 8, second person anyone except Bob or Rachel - 5 out of 7
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * \frac{5}{7}) = \frac{14}{56} = \frac{7}{28}\)

I feel method mentioned by 'walker' has repeated cases - when bob is selected in 1st position in some cases and then in second position in other cases - because we need selection here not permutations. Please clarify my doubt

Example 2
reversal probability approach : 1 - Probability when out of 3 person, two persons are a couple.
Prob. of selecting any one = \(\frac{1}{10} \)
Prob. of selecting the partner of the one selected = \(\frac{1}{9} \)
Prob. that third person be anyone = 1
So, Probability that out of 3 person, two persons are a couple = \(\frac{1}{10} * \frac{1}{9} * 1 \) = \(\frac{1}{90} \)
So, Probability that 3 persons in group be anyone but a couple = 1 - \(\frac{1}{90} \) = \(\frac{89}{90} \)

I understand my answers are coming out to be wrong but please point out the mistakes in my approach. Appreciate your help a lot !
Thanks in advance
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Hello Bunuel souvik101990,

I'd like you to help me understand the significance of a concept.

Ex: A box contains 26 blocks, each listing a unique English alphabet. If 3 blocks are picked at a time, what is the probability that all 3 are vowels?

Approach 1: 5C3 / 26C3

Approach 2 : (5C1/26C1) * (4C1/25C1) * (3C1/24C1)

This approach gives the same answer as Approach 1 does; however, since there is no order mentioned why this approach works? What is the logical significance of this approach?

Approach 3 : (5C1 * 4C1 * 3C1 / 26C3)

What is the significance of this approach. Does it mean the same as approach 2 does?

For long I have struggled to understand this conceptual nuance.

Much appreciated,

Thanks
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jasonedward
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.
Show more


**********************************************************************************

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool = Probability of Donald NOT getting Progaine * Probability of Donald Getting Ropecia = 13/14 * 1/13 = 1/14.

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

Hope this helps ! :)
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Math: Probability 20 Oct 2021, 13:06
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Wow - between Walker and Bunuel, we have Math covered.
Great job!
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Hi folks,

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=12∗P52=10


I think I am missing something silly here. So, let's say I consider Bob & Rachel as 1 unit - with proper arrangement (meaning, BR (not RB)).
I have 4 items. No of possible arrangements is 4! = 24.

What's wrong?

If you are arranging 2 objects into 5 slots where 'order' matters (meaning ABCDE is different from BACDE), you do 5P2. But, then I get that you take 1/2 of it to arrive at 10. What is logically wrong with 4!?

I got a disappointing Q49 recently. Going for a Q50/51. Need to sort out basic stuff soon..
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shrgmat
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Math: Probability 20 Oct 2021, 13:25
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Thanks Bunuel. Yes, that makes sense. I don't have other options then. Must use 5P2 / 2.
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35wlBAB9081p
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Math: Probability 11 Feb 2022, 15:15
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walker

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).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)
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Thank you for great lesson walker!! I try to calculate opposite probability to problem example you have. I get right answer with combinatorial approach (1/3). I dont get right answer with probability approach.

My problem: If choosing 3 out of 10, what is probability of getting one couple who is married?

Probability of choosing any person: 10/10 = 1
Probability of choosing the wife: 1/9
Probability of choosing any person: 8/8 = 1

Total probability = 1 * 1/9 * 1 = 1/9

This is wrong. What is mistake in thinking here?
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