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Only Jim and not Jill

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by sanju09 » Sat Feb 21, 2009 5:22 am
3 people are to be selected out of 8 people. Out of the three, Only Jim and not Jill needs to be selected. What is the probability of this?
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by Alpturk » Sat Feb 21, 2009 8:23 am
answer must be 5/56

solution: 5! / 1!.4! = 5 if Jim and Jill are selected.

however if there is no condition about Jim and Jill 8! / 3!.5! = 56

then the answer equal to 5/56
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by Alpturk » Sat Feb 21, 2009 8:36 am
i tried to correct my previous answer.
answer, i think that, might be 15/64
?
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by tkarthi4u » Sat Feb 21, 2009 8:33 pm
Total number of ways 3 is selected out of 8 people = 8C3 ways = 48

Total no of ways Jim is Selected and Jill not = 1*6C2 = 15

The probability of this = 15/48.

Please let me know if i am wrong.
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by sanju09 » Mon Feb 23, 2009 4:41 am
tkarthi4u wrote:Total number of ways 3 is selected out of 8 people = 8C3 ways = 48

Total no of ways Jim is Selected and Jill not = 1*6C2 = 15

The probability of this = 15/48.

Please let me know if i am wrong.
When Jim is already declared selected, and Jill is already declared as not in favorites, isn't this left as select any two among the remaining six with no favor given? Isn't that 100% possible? I am only asking, not answering it. :)
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by rsadana1 » Wed Feb 25, 2009 11:47 am
You may be missing a key point here.

Probability = number of desired outcomes / total number of outcomes.

Now in this case, the total number of outcomes pertains to the number of ways in which 3 people can be selected from 8 people. So this is your TOTAL OUTCOMES.

The situation that you have explained is the DESIRED OUTCOME in which Jim should be selected as 1 of the 3 people and Jill should not be in the selection pool for the remaining 2 vacancies. So we need to determine the total number of ways in which the desired outcome can be accomplished and then calculate the probability by plugging the values in the formula for probability.

Does this make things any clearer?
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by sanju09 » Thu Feb 26, 2009 1:34 am
rsadana1 wrote:You may be missing a key point here.

Probability = number of desired outcomes / total number of outcomes.

Now in this case, the total number of outcomes pertains to the number of ways in which 3 people can be selected from 8 people. So this is your TOTAL OUTCOMES.

The situation that you have explained is the DESIRED OUTCOME in which Jim should be selected as 1 of the 3 people and Jill should not be in the selection pool for the remaining 2 vacancies. So we need to determine the total number of ways in which the desired outcome can be accomplished and then calculate the probability by plugging the values in the formula for probability.

Does this make things any clearer?
B-) That's cool; but how would you rebuff my point, rsadana1?
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by rsadana1 » Thu Feb 26, 2009 7:05 am
"When Jim is already declared selected, and Jill is already declared as not in favorites, isn't this left as select any two among the remaining six with no favor given? Isn't that 100% possible?"

Yes it is 100% possible but as the desired outcome. This statement exactly represents the desired outcome. However, you need to know what the reference point of calculating the probability is - it is the case when any 3 people are to be selected from the group of 8 people. So in probability questions, you always need to understand two things independent of each other - first is the total outcomes and second is the desired outcomes.

Have I made things clear or do you still have doubts? :) [/b]
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by Alpturk » Fri Feb 27, 2009 11:00 am
sorry what is the answer of this question?
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by rsadana1 » Fri Feb 27, 2009 12:54 pm
15/48 is the answer.
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by sanju09 » Sat Feb 28, 2009 4:43 am
rsadana1 wrote:15/48 is the answer.
How is 8C3 = 48? Though I am still not convinced with your explanation :)
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by Alpturk » Sat Feb 28, 2009 5:38 am
in my opinion, the probablity of jim to be selected is 3/8 and the probabilty of not being selected jill is 5/8.
when we multiply that 3/8 and 5/8, it equals to 15/64.
you think that that is not logical?
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by sanju09 » Sat Feb 28, 2009 5:42 am
Alpturk wrote:in my opinion, the probablity of jim to be selected is 3/8 and the probabilty of not being selected jill is 5/8.
when we multiply that 3/8 and 5/8, it equals to 15/64.
you think that that is not logical?
Are there 3 Jims and 3 Jills in the said group of 8, my friend? :)
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by DanaJ » Sat Feb 28, 2009 9:05 am
The wording of your problem makes me wonder what you actually mean. On a first glace (and I'm probably not going to give it a second glance, since I HATE probabilities), you're looking for (the number of 3 person groups containing Jack but not Jill)/(total number of groups of 3 people)
The total number of groups will be 8C3 = 56.
Groups that contain Jack have "two open positions" for the other 6 people. There are 6C2 possible groups of two people from the pool of 6 left or 15.
This makes the solution to this problem (IMHO, of course, since I DESPISE probabilities) 15/56.
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by rsadana1 » Sat Feb 28, 2009 5:24 pm
oops my bad. you are right Sanju09, 8C3 cannot be 48...That was an oversight on my part. The correct answer should be 15/56 as pointed out by DanaJ since 8C3 is 56 and 6C2 is 15.

Sorry for misleading the answer in this post...:( But I still standby my explanation!
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