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A certain television show has 15 sponsors, including Company

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A certain television show has 15 sponsors, including Company [#permalink]

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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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Re: A certain television show has 15 sponsors, including Company [#permalink]

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New post 12 Feb 2012, 06:25
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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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Re: A certain television show has 15 sponsors, including Company [#permalink]

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New post 21 May 2014, 17:54
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Re: A certain television show has 15 sponsors, including Company [#permalink]

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New post 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
Probability.PNG [ 2.83 KiB | Viewed 1979 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
Probability2.PNG [ 2.22 KiB | Viewed 1976 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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Re: A certain television show has 15 sponsors, including Company [#permalink]

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New post 22 May 2014, 03:45
Expert's post
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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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a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A certain television show has 15 sponsors, including Company [#permalink]

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New post 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
Re: A certain television show has 15 sponsors, including Company   [#permalink] 21 Feb 2016, 10:31
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