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May 31st, 2011, 10:24 PM  #1 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Probability(Hard)
1.a)One letter is selected from each of the names: SIMMS,SMITH,THOMPSON.What is the probability that 2,and only 2 are the same? Answer:0.34 b)A candidate attempts a question to which 5 possible answer have been given,one of them correct.For any question,there is a probability of 1/3 that he knows the correct answer.If he does not know the correct answer he will mark one of the answer at random.He does,in fact,mark the correct answer.What is the probability that he knew the correct answer? Answer:5/7 I need help and explanation for the question.I don even know how to start solve the question.Thank you. 
June 1st, 2011, 05:46 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Probability(Hard) Hello, hoyy1kolko! Quote:
There is no formula for this problem. [color=beige]. . [/color]We must make an exhuastive list. We have:[color=beige]. . [/color] And there are five case of doubled letters . . . [color=beige]. . . . . . . [/color] [color=beige]. . . . . . . [/color]  
June 1st, 2011, 07:00 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Probability(Hard) Hello again, hoyy1kolko! Quote:
We want: the probability that he knew the answer, given that he marked the correct answer. [color=beige]. . [/color] [color=beige]. . [/color]  
June 1st, 2011, 07:42 AM  #4 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: Probability(Hard)
Your subject line is too cute and your 2nd problem is good fun, so I willl start with 2nd... The candidate knows 1/3 question's answer, that means he doesn't know 2/3 questions answer If he is attempting the question he knows then he will just pick the right answer in 1 way If he is attempting the question he doesn't know then he may pick the correct answer in 1/5 way or may pick the wrong answer in 4/5 ways Since the right answer is picked, we will calculate how many ways are there The total no. of ways for a right answer to be picked is 1/3*1+2/3*1/5 = 7/15 The no. of ways that he knows the answer is 1/3*1 = 1/3 Probability is (1/3)/(7/15) = 5/7 First problem... Let's say any letter picked from SIMMS is x, from SMITH is y, from THOMPSON is z your condition is x=y not= z or x not= y=z or x=z not= y x=y not= z those letters could be S,I and M for x and y and not for z so x can be S in 2/5 ways, y can be S in 1/5 ways and z can't be S in 7/8 ways = (2/5*1/5*7/, similarly you have to calculate the wayz for I and M that would be (2/5*1/5*7/+(1/5*1/5*8/+(2/5*1/5*7/ = 18/100 x not= y=z here the letters are S,M,T and H similarly as above you will get the total number 8/100 x=z not= y here the letters are S and M similarly as above you will get the total number 8/100 If you will sum up you will get 34/100=.34 
June 1st, 2011, 08:12 AM  #5 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: Probability(Hard)
Thank for both answer from soroban and rudrax,it helps me understand better and the explanation are interesting. I still don't understand on soroban's solution on question 1 (a) about probability that 2 letter are the same. And I don't understand how the diagram is formed. I need more explanation on it.Thank you. 
June 4th, 2011, 04:55 PM  #6  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
 

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