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# A certain television show has 15 sponsors, including Company

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Director
Joined: 07 Jun 2004
Posts: 605
Location: PA

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12 Feb 2012, 06:15
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Question Stats:

82% (01:22) correct 18% (00:58) wrong based on 52 sessions

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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 Xs advertisement will be one of the first two shown during the first commercial break?

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?
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Senior Manager
Status: You have to have the darkness for the dawn to come
Joined: 09 Nov 2012
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GMAT 1: 590 Q49 V20
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07 Feb 2017, 04:17
3
rxs0005 wrote:
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 Xs advertisement will be one of the first two shown during the first commercial break?

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?

let's the first show be x's show then the probability= 1/15 --------------------------1
consider the 2nd show is x's show then probability =(14/15)*(1/14)= 1/15---------------------2
(1/15)+(1/15) =2/15
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##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 47184

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12 Feb 2012, 06:25
1
2
rxs0005 wrote:
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 Xs advertisement will be one of the first two shown during the first commercial break?

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?

This is basically the same question as this one: a-medical-researcher-must-choose-one-of-14-patients-to-127396.html

The probability that Company X's advertisement to be one of the first two is simply 2/15 as there are 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 on second slot it shouldn't appear on first, so P(second slot)=14/15*1/14 and not 2/14 as you've written).

Hope it's clear.
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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:

Attachment:

Probability.PNG [ 2.83 KiB | Viewed 4313 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:

Attachment:

Probability2.PNG [ 2.22 KiB | Viewed 4312 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)
=$$\frac{(14+14)}{(15*14)}$$
=$$\frac{2}{15}$$
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Joined: 02 Sep 2009
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22 May 2014, 03:45
rxs0005 wrote:
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 Xs advertisement will be one of the first two shown during the first commercial break?

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?

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21 Feb 2016, 10:31
rxs0005 wrote:
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 Xs advertisement will be one of the first two shown during the first commercial break?

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?

Hi Bunuel,

Why this approach is wrong :
(1/15)(13/14)(12/13)(11/12) + (14/15)(1/14)(12/13)(11/12)

Thanks & regards,
Sunil01
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Joined: 02 Aug 2017
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19 Oct 2017, 16:08
Why is the probability for the second slot not 1/14? One advertiser has already filled slot 1 so there are 14 remaining, of which the company is one. The question states that each ad is from a different company.
Re: A certain television show has 15 sponsors, including Company &nbs [#permalink] 19 Oct 2017, 16:08
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