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 July 14th, 2011, 07:39 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability question(Interesting) A calculator generates one of the 8 consecutive digits 1,2,.....8 such that the probability of getting an odd digit is twice that of getting an even digit.State the probability of getting an odd digit.The calculator is used to generate 4 digit,each independent of the preceding ones,to form a 4-digit number(e.g if the digits 1,2,7,8 are generated in that order,the number formed is 127 Find the probability that a)the number formed is an odd number greater than 7000 b)the 3rd and 4th digit of the number formed are the same c)the 3rd and 4th digits of the number formed are the same,given that it is an odd number greater than 7000. I am having problem in understanding and solving the question.I need help.Please explain to me step by step of the question.Thank you. July 14th, 2011, 08:46 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability question(Interesting) Before starting you must solve the following problem: The probability of getting an odd digit is twice that of getting an even digit. The probabilities must sum to 1. Thus P(odd) = 2 * P(even) and P(odd)+P(even) = 1. Then, there are four odd digits to choose from, so P(1)=P(3)=P(5)=P(7)=P(odd)/4. Similarly, P(2)=P(4)=P(6)=P( =P(even)/4. Solve that problem first, and post your results and working. Now the digits are picked in order. For example, the probability of 7321 is P(7)*P(3)*P(2)*P(1). The probability of 73_1 (third digit can be anything) is P(7)*P(3)*P(1). a) An odd number greater than 7000 must start with 7 or 8, and end with an odd digit. b) The 1st digit can be anything, the 2nd digit can be anything, the 3rd digit can be anything, and then the 4th digit must be the same as the 3nd digit. So the probability is (chance 3rd digit is odd)*(chance 4th digit is then the same) + (chance 3rd digit is even)*(chance 4th digit is then the same) July 15th, 2011, 11:03 PM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question(Interesting) Sorry,i still don understand it. for part (a) i use _ _ _ _ > 7000 first digit,2P1 Second and third digit 8P2 4th digit 1,3,5,7 4P1 2P1 x 8P2 x 4P1 =448 448/8P4=4/15. I can't get the correct answer.the answer is 1/6. July 16th, 2011, 01:35 AM   #4
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Re: Probability question(Interesting)

Quote:
 Originally Posted by aswoods Before starting you must solve the following problem: The probability of getting an odd digit is twice that of getting an even digit. The probabilities must sum to 1. Thus P(odd) = 2 * P(even) and P(odd)+P(even) = 1. Then, there are four odd digits to choose from, so P(1)=P(3)=P(5)=P(7)=P(odd)/4. Similarly, P(2)=P(4)=P(6)=P( =P(even)/4. Solve that problem first, and post your results and working.
What did you get for P(odd) and P(even)?
What did you get for P(1), P(3), ... and for P(2), P(4), ... ? Tags probability, questioninteresting Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Saunak Algebra 2 September 6th, 2013 11:22 AM alpha_century Probability and Statistics 0 May 5th, 2013 12:22 PM superconduct Probability and Statistics 6 March 9th, 2012 07:38 AM hoyy1kolko Probability and Statistics 4 July 9th, 2011 01:36 PM Kevlar93 Algebra 12 December 10th, 2008 05:22 PM

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