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July 14th, 2011, 07:39 AM   #1
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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 127Find 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.
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July 14th, 2011, 08:46 AM   #2
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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)
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July 15th, 2011, 11:03 PM   #3
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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.
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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), ... ?
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