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April 5th, 2007, 12:51 PM  #1 
Newbie Joined: Apr 2007 Posts: 5 Thanks: 0  simple probability question
Hi every body could any one help me to solve this problem REGARDS 
April 5th, 2007, 06:39 PM  #2 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
For the first question, the probability of more than 3 errors is 1(probability of 3 or fewer errors)1  (1000 choose 3)*(1/1000)^3*(999/1000)^997  (1000 choose 2)*(1/1000)^2*(999/1000)^998  (1000 choose 1)*(1/1000)^1*(999/1000)^999  (1000 choose 0)*(999/1000)^1000 Try to figure out what this means, why it is true, and multiply it out. Feel free to ask questions. For the second, an input signal will reach the output iff all three switches are closed. This happens with a probability of p³. Also, if an input switch was received at the output, then all three switches must be closed. 
April 6th, 2007, 04:40 AM  #3 
Newbie Joined: Apr 2007 Posts: 5 Thanks: 0 
Thank you very much indeed my friend roadnottaken for your reply.Your explaination is very very clear and you have made me confudent about my self. I am still have a problem that i want to make sure about it. I wonder if you could help me again. I know that i am very curious, but excuse me please RAGEDS Your friend The_Master 
April 6th, 2007, 06:41 AM  #4 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Could you please type the questions rather than using such large images? Oops. I worked question three assuming the switches operated in series rather than in parallel. Since they work in parallel, the probability that an input signal is received is the opposite of the probability that all three switches are open. The actual probability that a signal is received is 1(1p)³ = 1(13p+3p²p³) = p³3p²+3p. The conditional probability that switch 1 was open, then, is (1p)/(p³3p²+3p). The conditional probability A given B is simply the probability of A divided by the probability of B. Since there is a 50% chance that any given component works and a 1.5^n probability that an input signal is received as output, the conditional probability is .5/(1.5^n). Say the probability of flipping a head is p. The probability that an odd number of tosses is required is p[1+(1p)²+(1p)^4+(1p)^6...] = p*∑_(n=0)^∞ (1p)^(2n) This is an infinite geometric series with first term p and common ratio (1p)²=p²2p+1, so, because 0<p<1, the sum is p/[1(p²2p+1)]=p/(p²+2p)=1/(2p). Don't worry too much if you don't understand the ∑ notation, but make sure you understand everything else, and ask questions if you don't. 
April 6th, 2007, 02:15 PM  #5 
Newbie Joined: Apr 2007 Posts: 5 Thanks: 0 
Thank you my friend. I cannot say any thing unless telling you that you are brilliant.... Frankly i still have some problems and i want to make sure about them. I'm very shy to say that i couldn't wirte the words in the below image because there are some sympols that i couldn't be able to write them. Any way this the question and i will be happy if you could help me dear. SORRY agin for this image REAGRDS You frind The_Master 

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