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Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 11:30

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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected? (A) 0.1

(B) 0.2

(C) 0.25

(D) 0.4

(E) 0.6 Ans: This means that P(M and N both selected) = (2/5) x (1/4) = 1/10

Another question:: Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

a)1/15 b)1/12 c)1/9 d)1/6 e)1/3

Ans: (1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15 Here why do we consider Both the options (Jose*Joshuna) and (Joshuna*jose) But in the first question we considerded probability just 2/5x1/4 and why not (2/5x1/4)(1/4x2/5) like we did in the second problem.Please explain the difference.. Please explain

Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 11:49

I believe the order in which Joshua and Jose appear the interview is taken into account, even though the question does not at any point spell that out clearly.

Normally such cases where the order in which the events are to happen is taken into consideration, the above principle is taken into account. In this case , (Jose*Joshuna) and (Joshuna*jose) was taken to take into account so that Jose went to the interview first and then Joshua, and for the second case, Joshua went first and then Jose. Although I am not entirely sure, if the above process should be applied to a question like the one mentioned, I'd definitely request you to confirm the answer from the source

Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 11:54

I don't know whether my answer will answer you question, but i'll try my best. 1st problem: Total possibilities: selecting two from the group of 5 people. as we r selecting and not arranging the possibilies will be 5C2 i.e. 10 required possibility: only 1 combination (Marnie and Noomi or Noomi and marnie)..arragement/order doesn't matter. so answer = (required possibilitt)/(total ppssibility) i.e 1/10 i.e. 0.1

2nd problem is same, just that number of people are more. total possible: again we r just selecting, not arranging, so 6C2 i.e 15 Required: only 1 combination so answer=1/15

hope everything is now clear.

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Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 12:30

my question is why in first question probability is calculated and for the second questions they are doing the same thing twice why? i am expecting an answer to be from probability point of view oly
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Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 14:00

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skamal7 wrote:

my question is why in first question probability is calculated and for the second questions they are doing the same thing twice why? i am expecting an answer to be from probability point of view oly

As I have mentioned in my post above, the reason the probabilities are added is to take into account the order in which the interview was taken i.e. if the order to the interview was {Jose, Joshua} or {Joshua,Jose}. Lemme walk you thru the answer.

The probability of randomly choosing {Jose} from the set of 6 workers is clearly 1/6. Now after already have chosen one, the probability to choose the {Joshua} or any other worker would be 1/5 So the net probability of the interview order{Jose, Joshua} is 1/5*1/6 = 1/30. (Let it be P1)

Now, since the assumption to be made here is the interview order matters i.e. the interviews are one-on-one. The other possible order would be {Joshua, Jose} The probability to happening of the above order is also 1/30 (Let it be P2). Hence the final Probability of both the possible cases is P1 + P2 = 1/30 + 1/ 30 = 1/15.

But, in the first question, 2 people are supposed to go to a conference from a group of 5. The order here i.e. {Marnie, Noomi} or {Noomi,Marnie} does not matter. Hence the probability of choosing any random person from either (Marie or Noomi) in a group of 5 = 2/5 And then, the probability of choosing the next person from (Marie or Noomi) in a group of 4 = 1/4 Final probability = 2/5 * 1/4 = 1/10

Hope this satisfies your query! Would appreciate it if the mods had a look at it and correct me if I am wrong

Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 14:51

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Lets see if I can help here:

First of all a bit of theory:

Combinations : the order is NOT relevant What do I mean? Changing the order DOES NOT result in a new arrangement

Example Combinations:"select a team of 5 from a group of 10". In this case the team remains the same and is not affected by the order I choose the players To make it clear under there is an example of Perutaions is which the order IS important I don't wanna confuse you but I think it's important to know the difference Example Premutation: the very classic "How many ways can 5 students be arranged in a row of 10"? In this case the order is important

Now lets see the formulas N= numb items, K=spots Tot combinations= \(\frac{n!}{k!(n-k)!}\)

Now down to the questions: 1)From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected? (A) 0.1 (B) 0.2 (C) 0.25 (D) 0.4 (E) 0.6

The order is not important (M and N = N and M at the conference) in this case => Comb Tot cases=\(\frac{5!}{3!2!}=10\) case "good" M,N (or N,M order Not imp) so \(\frac{1}{10}\)

2)Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen? a)1/15 b)1/12 c)1/9 d)1/6 e)1/3

The order is not important => Comb Tot cases=\(\frac{6!}{2!4!}=15\) case good Josh,Jose (or Jose,Josh order Not imp) so \(\frac{1}{15}\)

I suggest you to use this approach to solve questions like those However from a probability point of view:

This means that P(M and N both selected) = (2/5) x (1/4) = 1/10 2/5 Means one of the two (2 out of 5), 1/4 means the other

(1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15 Here 1/6 and 1/5 are singular and for this must be taken twice. please note the difference In the first we have (one of the two)*(the other). This already contains the "other way" In the second we have (Josh)*(Jose), but this does not contain the "other way". That's why also (Jose)*(Josh) must be taken into consideration If we solve the second as the first we would get: 2/6*1/5=1/15 => same approach as one (one of the two)*(the other)

(Tip: use comb are more clear )

Sorry for the lenght of the post but I felt like clarity>synthesis here Hope it's clear. Let me know
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Re: Joshua and Jose work at an auto repai [#permalink]

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14 May 2013, 21:50

Zarollu, You play a vital part in my preperation man..

Please correct me whether am i right if the question is stated as From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 3 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie ,Noomi and lee are both selected?

Here the answer would be 5x4x3/3x2 =10

ase "good" M,N,L (or N,M,L order Not imp) so ans is 1/10

Is my understanding correct?
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"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: Joshua and Jose work at an auto repai [#permalink]

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15 May 2013, 01:11

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skamal7 wrote:

Zarollu, You play a vital part in my preperation man..

Please correct me whether am i right if the question is stated as From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 3 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie ,Noomi and lee are both selected?

Here the answer would be 5x4x3/3x2 =10

ase "good" M,N,L (or N,M,L order Not imp) so ans is 1/10

Is my understanding correct?

As you say choose 3 out of 5 gives us 10 arrang. These are J K L, J K M, J K N, J L M, J L N, J M N, K L M, K L N, K M N, L M N

Remeber that the order does not matter. In how many cases do we find M N L? Just one ( the last one).

You're correct. If you have more doubts let me know
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