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Probability - Conceptual

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by JakilD » Wed Oct 12, 2011 3:23 am
If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

IMO - [spoiler]~ B ~[/spoiler]

Explanation - [spoiler]~ 18C3 / 20C5 ~[/spoiler]
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by mukgera » Wed Oct 12, 2011 3:36 am
Correct.
JakilD wrote:If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

IMO - [spoiler]~ B ~[/spoiler]

Explanation - [spoiler]~ 18C3 / 20C5 ~[/spoiler]
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by sohrabkalra » Wed Oct 12, 2011 3:45 am
Total ways of selecting 5 out of 20 = 20C5
Desirable = Since 5 and 15 both have to be selected, we are left with choices of selecting 3 out of remaining 18 = 18C3

So probability = 18C3/20C5 = 1/19 in my opinion
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by Proleefeek111 » Wed Oct 12, 2011 4:13 am
JakilD wrote:If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

IMO - [spoiler]~ B ~[/spoiler]


5 nos can be selected from 20 in 20C5 ways.

No of ways in which both 5 and 15 are selected = 18C3 <-- **With both 5 and 15 selected we have to fill 3 slots from 18 choices**

P(both 5 and 15 are selected) = 18C3 / 20C5 = 1/19

IMO B.
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by GmatMathPro » Wed Oct 12, 2011 6:22 am
This is for someone who PMed me about a way to do it without combinatorics...

1/10 chance of picking 5 or 15 on the first pick. 1/19 chance of picking the other one on the 2nd pick. 100% chance of picking anything else on the next 3 picks. 10 ways to select where the 5 and 15 go,

(1/10)(1/19)(10)=[spoiler]1/19[/spoiler]
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by knight247 » Wed Oct 12, 2011 6:59 am
GmatMathPro wrote:10 ways to select where the 5 and 15 go,

(1/10)(1/19)(10)=[spoiler]1/19[/spoiler]
Ok Pete,
Am gonna ask u a silly question now.lol. Assume our sequence of events is 5 15 X Y Z How is it that they can be arranged in 10 Ways? I guess u did a 5C2. I really don't understand how a 5C2 can be used here.
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by GmatMathPro » Wed Oct 12, 2011 8:48 am
There are 5 slots. Two of them have to be 5 and 15. Three of them have to be anything else. The way I set it up, I said that the first slot has a 1/10 slot of being either 5 or 15, and if that happens, the 2nd slot has a 1/19 chance of being the other number. So this includes the probability of getting 5 then 15 or 15 then 5. But of course they don't have to be in the first and second slots, they could be in any of the 5 slots. so we have to choose which two slots they will be in or 5C2.

So, we choose where the 5 and 15 go, but the way I set up the problem already includes the possibility of different orders of 5 and 15. And we don't have to worry about the different orders of the "anything else" slots because the calculation already includes all the different orders of those slots. For example, 5 15 7 8 9 and 5 15 9 8 7 are both already considered because when we say the three slots can be ANYTHING else, the last 3 could go 7 8 9 just as easily as 9 8 7.
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