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12 Feb 2012, 06:15
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Difficulty:
45%
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Question Stats:
68% (01:41) correct
32%
(02:11)
wrong
based on 656
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A certain television show has 15 sponsors, including Company X, each of which has produced a 30-second advertisement to be televised during the show. The first commercial break will consist of 4 of these 30-second advertisements, each of which will represent a different sponsor. What is the probability that Company Xs advertisement will be one of the first two shown during the first commercial break?
A. 1/15
B. 19/210
C. 2/15
D. 1/5
E. 4/15
A. 1/15
B. 19/210
C. 2/15
D. 1/5
E. 4/15
The answer to this is 2/15
My approach was
it could appear on first slot which is 1/15 or second slot 2/14 so the final probability is
1/15 + 2 / 14 = 19/ 210 why is this wrong?
My approach was
it could appear on first slot which is 1/15 or second slot 2/14 so the final probability is
1/15 + 2 / 14 = 19/ 210 why is this wrong?
12 Feb 2012, 06:25
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rxs0005
This is basically the same question as this one: https://gmatclub.com/forum/a-medical-re ... 96.htmlThe probability that Company X's advertisement will be one of the first two is simply 2/15, as there are a total of 15 slots.
Or another way: P(first slot) + P(second slot) = 1/15 + 14/15 * 1/14 = 1/15 + 1/15 = 2/15. (To appear in the second slot, it shouldn't appear in the first, so P(second slot) = 14/15 * 1/14 and not 2/14 as you've written).
Hope it's clear.
22 May 2014, 03:38
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The quickest way to solve this question is indeed to simply compute the probability that X gets the 1st slot (\(\frac{1}{15}\)) and the probability that X gets the 2nd slot (again \(\frac{1}{15}\)) and then add the two.
However, it may be useful for at least a few students to know another method to solve this problem:
Probability.PNG [ 2.83 KiB | Viewed 17759 times ]
The number of ways in which the first ad slot can be filled = 15
Number of ways in which the second ad slot can be filled (after the first one has already been filled) = 14
So, the number of ways in which both the slots can be filled together = 15*14
Now, let's consider the case when X gets Slot 1:
Probability2.PNG [ 2.22 KiB | Viewed 17633 times ]
The number of ways now in which the second slot can be filled = 14.
So, total number of ways in which X can get Slot 1 = 14
Similarly, the total number of ways in which X can get Slot 2 = 14
Probability of X getting Slot 1 or Slot 2 = (Number of ways in which X can get Slot 1 or Slot 2)/ (Total number of ways in which the 2 slots can be allocated)
=\(\frac{(14+14)}{(15*14)}\)
=\(\frac{2}{15}\)
However, it may be useful for at least a few students to know another method to solve this problem:
Attachment:
Probability.PNG [ 2.83 KiB | Viewed 17759 times ]
Number of ways in which the second ad slot can be filled (after the first one has already been filled) = 14
So, the number of ways in which both the slots can be filled together = 15*14
Now, let's consider the case when X gets Slot 1:
Attachment:
Probability2.PNG [ 2.22 KiB | Viewed 17633 times ]
The number of ways now in which the second slot can be filled = 14.
So, total number of ways in which X can get Slot 1 = 14
Similarly, the total number of ways in which X can get Slot 2 = 14
Probability of X getting Slot 1 or Slot 2 = (Number of ways in which X can get Slot 1 or Slot 2)/ (Total number of ways in which the 2 slots can be allocated)
=\(\frac{(14+14)}{(15*14)}\)
=\(\frac{2}{15}\)
