Events & Promotions
|
|
GMAT 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.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
28 Mar 2011, 09:39
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
There are 3 transmitters that are transmitting A and B symbols. Probability that symbol was sent from each source are
\(P(s1)=\frac{5}{10}\)
\(P(s2)=\frac{3}{10}\)
\(P(s3)=\frac{2}{10}\)
Probability of sent symbols on seperate source are
\(P(A1)=\frac{3}{7}\) , \(P(B1)=\frac{4}{7}\)
\(P(A2)=\frac{2}{3}\) , \(P(B2)=\frac{1}{3}\)
\(P(A3)=\frac{3}{5}\) , \(P(B3)=\frac{2}{5}\)
1. What is probability that reciever recieved a symbol B?
2. Reciever recieved 2 A symbols in a row. What is the probability that source 3 transmitted them?
\(P(s1)=\frac{5}{10}\)
\(P(s2)=\frac{3}{10}\)
\(P(s3)=\frac{2}{10}\)
Probability of sent symbols on seperate source are
\(P(A1)=\frac{3}{7}\) , \(P(B1)=\frac{4}{7}\)
\(P(A2)=\frac{2}{3}\) , \(P(B2)=\frac{1}{3}\)
\(P(A3)=\frac{3}{5}\) , \(P(B3)=\frac{2}{5}\)
1. What is probability that reciever recieved a symbol B?
2. Reciever recieved 2 A symbols in a row. What is the probability that source 3 transmitted them?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
28 Mar 2011, 10:33
Kudos
Bookmarks
1. Probability of receiving B is just the sum of probabilities of receiving B from each of transmiteres, which is equal to sum of product of probability of receiving a signal from each transmiter and probability of sending B by this transmiter:
\(\frac{5}{10}*\frac{4}{7}+\frac{3}{10}*\frac{1}{3}+\frac{2}{10}*\frac{2}{5}=\frac{2}{7}+\frac{1}{10}+\frac{2}{25}=\frac{175}{350}+\frac{35}{350}+\frac{28}{350}=\frac{238}{350}=\frac{119}{175}=\frac{17}{25}\)
\(\frac{5}{10}*\frac{4}{7}+\frac{3}{10}*\frac{1}{3}+\frac{2}{10}*\frac{2}{5}=\frac{2}{7}+\frac{1}{10}+\frac{2}{25}=\frac{175}{350}+\frac{35}{350}+\frac{28}{350}=\frac{238}{350}=\frac{119}{175}=\frac{17}{25}\)
28 Mar 2011, 10:53
Kudos
Bookmarks
2.By Bayes formaula the probability that 2A are produced by 3 transmiter equals to the probability that 3 transmiter made 2 A multiplied by the probability that 2 signals were made by 3 transmiter divided by the probability that we recieved 2 A
Probability that 3rd transmiter made 2 A is \(\frac{3}{5}*\frac{3}{5}=\frac{9}{25}\)
Probability that 2 signals were made by 3rd transmitter is \(\frac{2}{10}*\frac{2}{10}=\frac{1}{25}\)
Probability to receive 2A is equal to the sum of product of corresponding probabilities of making A by some particular transmitter and probabilities that this transmitter sent a signal:
\(\frac{3}{5}*\frac{3}{5}*\frac{2}{10}*\frac{2}{10}+\frac{2}{3}*\frac{2}{3}*\frac{3}{10}*\frac{3}{10}+\frac{3}{7}*\frac{3}{7}*\frac{5}{10}*\frac{5}{10}+2*\frac{3}{5}*\frac{2}{3}*\frac{2}{10}*\frac{3}{10}+2*\frac{3}{5}*\frac{3}{7}*\frac{2}{10}*\frac{5}{10}+2*\frac{2}{3}*\frac{3}{7}*\frac{3}{10}*\frac{5}{10}=\frac{9}{625}+\frac{1}{25}+\frac{9}{196}+\frac{6}{125}+\frac{9}{175}+\frac{3}{35}=\frac{34969}{122500}\)
Finally, we have
\(\frac{\frac{9}{25}*\frac{1}{25}}{\frac{34969}{122500}}=\frac{9*196}{34969}=\frac{1764}{34969}\)
Actually, there should be something wrong in my solution)))
Probability that 3rd transmiter made 2 A is \(\frac{3}{5}*\frac{3}{5}=\frac{9}{25}\)
Probability that 2 signals were made by 3rd transmitter is \(\frac{2}{10}*\frac{2}{10}=\frac{1}{25}\)
Probability to receive 2A is equal to the sum of product of corresponding probabilities of making A by some particular transmitter and probabilities that this transmitter sent a signal:
\(\frac{3}{5}*\frac{3}{5}*\frac{2}{10}*\frac{2}{10}+\frac{2}{3}*\frac{2}{3}*\frac{3}{10}*\frac{3}{10}+\frac{3}{7}*\frac{3}{7}*\frac{5}{10}*\frac{5}{10}+2*\frac{3}{5}*\frac{2}{3}*\frac{2}{10}*\frac{3}{10}+2*\frac{3}{5}*\frac{3}{7}*\frac{2}{10}*\frac{5}{10}+2*\frac{2}{3}*\frac{3}{7}*\frac{3}{10}*\frac{5}{10}=\frac{9}{625}+\frac{1}{25}+\frac{9}{196}+\frac{6}{125}+\frac{9}{175}+\frac{3}{35}=\frac{34969}{122500}\)
Finally, we have
\(\frac{\frac{9}{25}*\frac{1}{25}}{\frac{34969}{122500}}=\frac{9*196}{34969}=\frac{1764}{34969}\)
Actually, there should be something wrong in my solution)))
