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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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
adding 1 and 2
(1/15)+(1/15) =2/15

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:


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:



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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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

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.

Could some one advise if this method is correct.
X probability of being in the first round of ads= 4/15
X probability of being in the first 2 slots of the first 4 ads= 2/4
X probability of being in the first round of ads AND first 2 slots of first 4 ads= 4/15*2/4= 2/15

Any advise would be appreciated.

Why is the following incorrect: (14C1 * 1C1 + 14C1 * 1C1)/(15C2)?
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there are two ways to solve it. I m stuck with the second way of solving it.
first it could be shown first, or second. There are fifteen possible positions in which it could be shown: anywhere from first to last. Thus, the probability is the number of desired outcomes over the number of possible outcomes: 2/15.

second However, this method is not giving me the right answer:
probability of coming in the first place 1/15. probability for coming on the second place would be 1/14( As the first advertisement will not be repeated so the no. of options left will be 14 so it should be 1/14). and total probability would be (1/15)+(1/14)

Where i m going wrong in my second approach

Probability of X being in second place will not be 1/14 . This case involves dependent events , so you need to account for the probability of first place as well.

Probability that X will be on second place = 14/15 * 1/14 = 1/15 .

Hope that clarifies!

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