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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).

1 Answer

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Answer: (a) 0.28

(b) 0.89

(c) 0.22

Explanation:

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


\\The first digit can be any of the 9 digits
\\


\\The second digit can be any of the remaining 8 digits


\\The third digit can be any of the remaining 7 digits


\\The forth digit can be any of the remaining 6 digits


\\Therefore arrangement of the 4-digit number implies


\\Arrangement = 9 x 8 x 7 x 6


\\= 3024


\\It can also be written as: 9P4, which is also 3024.


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


\\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)


\\The first digit can be any of the 7 digits


\\The second digit can be any of the remaining 6 digits


\\The third digit can be any of the remaining 5 digits


\\The forth can be any of the remaining 4 digits


\\Therefore, arrangement = 7 x 6 x 5 x 4 = 840


\\(a) p( number ≥3500) = number≥3500/3024


\\= 840/3024


\\= 0.28


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


\\Once the first digit has been filled , we have eight digits remaining


\\Second digit can be any of these eight digits


\\Third digit can be any of the remaining seven digits


\\Forth digit can be any of the remaining six digits


\\Therefore , Arrangement = 4 x 8 x 7 x 6 = 1344


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


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


\\Once the last digit has been filled , we have eight digits remaining


\\First digit can be any of these eight digits


\\Second digit can be any of the remaining seven digits


\\Third digit can be any of the remaining six digits


\\Therefore , Arrangement = 4 x 8 x 7 x 6 = 1344


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


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


\\2688/3024


\\= 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

User Mhh Lecker
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