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Of the 4-digit numbers using the digits 1 to 9 without repeats, I pick one at random. Determine the following probabilities. (a) Prob (my number is ≥ 3500). (b) Prob (my number begins or ends with an even digit). 4 points (c) Prob (my number has exactly 2 even digits and is an even number).

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Answer: (a) 0.28

              (b) 0.89

              (c) 0.22

Step-by-step explanation:

Since repetition is not allowed, and we are forming 4 digits number from 1,2 ,…9

[tex]\\[/tex]The first digit can be any of the 9 digits [tex]\\[/tex]

[tex]\\[/tex]The second digit can be any of the remaining 8 digits

[tex]\\[/tex]The third digit can be any of the remaining 7 digits

[tex]\\[/tex]The forth digit can be any of the remaining 6 digits  

[tex]\\[/tex]Therefore arrangement of the 4-digit number implies

[tex]\\[/tex]Arrangement = 9 x 8 x 7 x 6  

[tex]\\[/tex]= 3024

[tex]\\[/tex]It can also be written as: 9P4, which is also 3024.  

[tex]\\[/tex]The next thing is to get 4 –digit number ≥3500 that can be formed from the digits 1 , 2 …9

[tex]\\[/tex]For the 4 – digits to be ≥ 3500, then the first digit will be either of the following set { 3,4,5,6,7,8,9)

[tex]\\[/tex]The first digit can be any of the 7 digits

[tex]\\[/tex]The second digit can be any of the remaining 6 digits

[tex]\\[/tex]The third digit can be any of the remaining 5 digits  

[tex]\\[/tex]The forth can be any of the remaining 4 digits

[tex]\\[/tex]Therefore, arrangement = 7 x 6 x 5 x 4 = 840

[tex]\\[/tex](a) p( number ≥3500) = number≥3500/3024

[tex]\\[/tex]= 840/3024

[tex]\\[/tex]= 0.28

[tex]\\[/tex](b) If the first digit must be even, the it must be one of the following set { 2,4,6,8}

[tex]\\[/tex]Once the first digit has been filled , we have eight digits remaining

[tex]\\[/tex]Second digit can be any of these eight digits

[tex]\\[/tex]Third digit can be any of the remaining seven digits

[tex]\\[/tex]Forth digit can be any of the remaining six digits

[tex]\\[/tex]Therefore , Arrangement = 4 x 8 x 7 x 6 = 1344

[tex]\\[/tex]P( number begin with even digit) = 1344/3024

[tex]\\[/tex]Also, If the last digit must be even, then it must be one of the following set { 2,4,6,8}

[tex]\\[/tex]Once the last digit has been filled , we have eight digits remaining

[tex]\\[/tex]First digit can be any of these eight digits

[tex]\\[/tex]Second digit can be any of the remaining seven digits

[tex]\\[/tex]Third digit can be any of the remaining six digits

[tex]\\[/tex]Therefore , Arrangement = 4 x 8 x 7 x 6 = 1344

[tex]\\[/tex]P( number ends with even digit) = 1344/3024

[tex]\\[/tex]P(number begins or end with even digits ) = p( number begin with even digit) + p( number ends with even digit) = 1344/3024 + 1344/3024

[tex]\\[/tex]2688/3024

[tex]\\[/tex]= 0.89

(c) To get exactly two even number  

Since there are four even numbers, it could be the combination of any 2

Therefore , number of exactly 2 even numbers = 2 x 8 x 7 x 6 = 672

Therefore p( exactly 2 even number ) = 672/3024

     = 0.2

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