By a process to be described later, we will randomly form a 7-digit number where
ID: 3128340 • Letter: B
Question
By a process to be described later, we will randomly form a 7-digit number where o the digit in the millions place (leading digit) cannot be zero. . all 7 digits must be different. Consider the sample space S consisting of all possible 7-digit numbers satisfying the above constraints: All 7-igit numbers satisfying leading digit cannot be zero end in zero and those 7-digit numbers that do not end in zero Sall cigits different The sample space above is partitioned (divided or separated) into the 7-digit numbers that All 7-digit mmbers satisfying all digits different last dgitis zero All 7-digit numbers sati sfying: all digits different leading and last digit not zero Since the sample space contains literally thousands of elements, it is unreasonable to list them all. It shall be your task to count them: COUNTING ELEMENTS IN THE SAMPLE SPACE (8pts: 1,2,2,2,1) a) Total number of elements in the sample space Use the multiplication rule to count the 2 partitions of the sample space, S Number of elements in the sample space that end in 0 Use the multiplication rule, noting that the last digit must be zero. This is how many elements are in the upper part of the partitioned sample space. Number of elements in the sample space that do not end in 0 Use the multiplication rule, noting that the last digit cannot be 0. This is how many elements are in the lower part of the partitioned sample space. Number of elements in sample space that are odd: Use the multiplication rule, noting that the digit in the ones place must be odd. Number of elements in sample space Deduce from knowing how many that are even: are odd and how many there are total.Explanation / Answer
1.a) Total no. of numbers in the sample space ,
The millions place can be filled by any of 1,2,3,4,5,6,7,8 and 9. So, we have 9 options.
The next position can be filled by any of 0 to 9 except the one in the first place so there are 9 different ways for each of the number selected in the first place.The next place can be filled by rest of 8 numbers the next place again can be filled by 7 place ..continuing this way we have,
9*9*8*7*6*5*4 = 9*9!/3! ways .
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(b) No.of elements ( 7 digited ) ending with 0 is,
9*9*8*7*6*5*1 ( lastly 1 for onle 0 is to be chosen ) = 9*9!/4! .
No. of elements ( 7 digited ) ending not with 0 is,
9*9*8*7*6*5*3 ( last digit can be anyone of the rest 4 except 0 ) = 9*3*9!/4!
No. of elements that are Odd is,
the last number can be any of 1,3,5,7,9 so there are 5 options. Then the millions place can be filled by rest of the 8 numbers, the next one also can be filled by 8 ( this time including 0 ) ways..and so on.
So the total no. of odds becomes, 5*(8*8*7*6*5*4) .
now, 5*(8*8*7*6*5*4 ) = 5*8*8!/3!
The no. of numbers those are even is,
Since every number is either an Odd or an Even so total no. of even numbers is ,
9*9!/3! - 5*8*(8!/3!)
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The probability that any specific number does not end in 0,
(9*3*9!/4!) / (9*9!/3!) = 18/24 = 3/4 .
The probability that any specific number does end in 0,
(9*9!/4!) / (9*9!/3!) = 1/4 .
The probability that a number chosen is odd,
(5*8*8!/3!)/(9*9!/3!) = 40/81 .
The probability that a number chosen is even,
1 - 40/81 = 41/81 .
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