GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Beginning on Monday, once each day, Hector auditions for a role as a m
[#permalink]
11 Feb 2022, 09:31
11 Feb 2022, 09:31
Expert Reply
1
Bookmarks
00:00
Question Stats:
40% (01:52) correct
60% (02:00) wrong based on 5 sessions
Hide Show timer Statistics
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
A) \(\frac{1}{3}\)
B) \(\frac{5}{9}\)
C) \(\frac{2}{3}\)
D) \(\frac{19}{27}\)
E) 1
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Beginning on Monday, once each day, Hector auditions for a role as a m
[#permalink]
11 Feb 2022, 10:40
11 Feb 2022, 10:40
1
Carcass 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
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
gmatclubot
Re: Beginning on Monday, once each day, Hector auditions for a role as a m [#permalink]
11 Feb 2022, 10:40
Moderators:
