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A certain band will perform 15 different songs in random ord
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28 Dec 2019, 13:16
28 Dec 2019, 13:16
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A certain band will perform 15 different songs in random order, and no song will be performed twice. If 9 of the songs are new, what is the probability that the first 2 songs that the band performs will both be new?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{12}{35}\)
Re: A certain band will perform 15 different songs in random ord
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29 Dec 2019, 01:00
29 Dec 2019, 01:00
1
We have to select two songs such that both of them are new.
Total new songs available = 9
Probability = \(\frac{9C2}{15C2}\) = \(\frac{9*8}{15*14}\)
Probability = \(\frac{12}{35}\)
Total new songs available = 9
Probability = \(\frac{9C2}{15C2}\) = \(\frac{9*8}{15*14}\)
Probability = \(\frac{12}{35}\)
Carcass wrote:
A certain band will perform 15 different songs in random order, and no song will be performed twice. If 9 of the songs are new, what is the probability that the first 2 songs that the band performs will both be new?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{12}{35}\)
Re: A certain band will perform 15 different songs in random ord
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09 Jan 2020, 01:22
09 Jan 2020, 01:22
1
Carcass wrote:
A certain band will perform 15 different songs in random order, and no song will be performed twice. If 9 of the songs are new, what is the probability that the first 2 songs that the band performs will both be new?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{12}{35}\)
First and foremost the question is about arrangement and not selection
The Random experiment is about arranging 15 distinct songs which is 15!
The event is about arranging 2 new songs at the start (9P2) and then the remaining 13 songs (13!)
[9P2 * 13!][/15!]
= 12/35
