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consider all of the 4 digit numbers that can be made from the digits 0 to 9 (assume that the numbers cannot start with 0) . What is the probability of choosing a random number from this group that is less than or equal to 8000? Enter a fraction or round your answer to 4 decimal places, if necessary.

Sagot :

First, we need to determine the total amount of numbers fulfilling the conditions:

- 4 digits

- Not starting with 0

For the first digit, we have then 9 possible numbers: 1, 2, 3, 4, 5, 6, 7, 8 and 9.

For the second, third and fourth, we have 10 possible numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Then, to determine the amount of numbers available we just need to multiply the possibilities for each digit:

[tex]9\cdot10\cdot10\cdot10=9000[/tex]

Then, randomly choosing one of the given numbers, we have 9000 possible outcomes. Those will be numbers from 1000 to 9999.

Now we just need to determine how many numbers among those 9000 are lower than or equal to 8000.

As the numbers start in 1000, we have 7001 cases where the randomly selected number is lower than or equal to 8000.

We obtain 7001 since 8000 - 1000 = 7000 but we need to consider also the number 1000.

The probability will be then:

[tex]\frac{7001}{9000}\approx0.7779[/tex]

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