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Beginning on Monday, once each day, Hector auditions for a role as a m

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Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink] New post   25 Nov 2021, 07:05
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Beginning on Monday, once each day, Hector auditions for a role as a movie extra, and at each audition, he has a 1/3 probability of being hired. If he stops auditioning once he is hired for the first time, what is the probability he is hired before the end of Wednesday?

A) \(\frac{1}{3}\)

B) \(\frac{5}{9}\)

C) \(\frac{2}{3}\)

D) \(\frac{19}{27}\)

E) 1
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Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink] New post   25 Nov 2021, 07:17
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sjuniv32 wrote:
Beginning on Monday, once each day, Hector auditions for a role as a movie extra, and at each audition, he has a 1/3 probability of being hired. If he stops auditioning once he is hired for the first time, what is the probability he is hired before the end of Wednesday?

A) \(\frac{1}{3}\)

B) \(\frac{5}{9}\)

C) \(\frac{2}{3}\)

D) \(\frac{19}{27}\)

E) 1


Note: If P(Hector is hired) = 1/3, then P(Hector is NOT hired) = 2/3

If Hector is hired before the end of Wednesday, and that means Hector was hired on Monday OR on Tuesday OR on Wednesday.
Rather than consider all three possible cases, we can save some time by using the complement.
That is: P(hired before end of Wednesday) = 1 - P(NOT hired before end of Wednesday)

P(NOT hired before end of Wednesday)
P(NOT hired before end of Wednesday) = P(NOT hired on Monday AND NOT hired on Tuesday AND NOT hired on Wednesday)
= P(NOT hired on Monday) x P(NOT hired on Tuesday) x P(NOT hired on Wednesday)
= 2/3 x 2/3 x 2/3
= 8/27

So, P(hired before end of Wednesday) = 1 - 8/27
= 27/27 - 8/27
= 19/27

Answer: D
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Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink] New post   27 Nov 2021, 14:54
sjuniv32 wrote:
Beginning on Monday, once each day, Hector auditions for a role as a movie extra, and at each audition, he has a 1/3 probability of being hired. If he stops auditioning once he is hired for the first time, what is the probability he is hired before the end of Wednesday?

A) \(\frac{1}{3}\)

B) \(\frac{5}{9}\)

C) \(\frac{2}{3}\)

D) \(\frac{19}{27}\)

E) 1


Probability of getting hired = 1/3 so Probability of not getting hired = 2/3

Probability that Hector gets hired by Wednesday = P(Hired on Monday) + P(Hired on Tuesday) + P(Hired on Wednesday)
=> P(Hired by Wednesday) = P(Hired on Monday) + P(Not Hired on Monday) x P(Hired on Tuesday) + P(Not Hired on Monday) x P(Not Hired on Tuesday) x P(Hired on Wednesday)
=> P(Hired by Wednesday) = 1/3 + 2/3 x 1/3 + 2/3 x 2/3 x 1/3
=> P(Hired by Wednesday) = 19/27

Option D
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Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink] New post   22 Aug 2023, 10:30
Beginning on Monday, once each day, Hector auditions for a role as a movie extra, and at each audition, he has a 1/3 probability of being hired. If he stops auditioning once he is hired for the first time, what is the probability he is hired before the end of Wednesday?

Probability Hector is hired before the end of Wednesday = Probability Hector is hired on Monday + Probability Hector is hired on Tuesday+ Probability Hector is hired on Wednesday = (1/3) + (1/3)(2/3)+ (2/3)(2/3)(1/3) = (1/3) + (2/9)+ (4/27) = (9/27) + (6/27)+ (4/27) = 19/27

Hence D
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Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink] New post   10 Nov 2023, 21:27
hired = 1/3, not hired = 2/3

porb. of not getting hired all three days = 8/27

therefore, prob of getting hired before Wednesday = 1 - 8/27 = 19/27
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Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink]
10 Nov 2023, 21:27
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