Randomly pick a positive integer with exactly four decimal digits, that is, a positive integer between 1000 and 9999 inclusive. What is the probability that it is following properties: (a) is divisible by 5 but not by 7. equation editorEquation Editor (b) has distinct digits. equation editorEquation Editor (c) even. equation editorEquation Editor (d) is divisible by 5.

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Answer:

a)0.1714 b)0.504 c)0.5 d)0.2

Step-by-step explanation:

Total numbers between 1000 and 9999 (both inclusive)= 9000.

Probability=[tex]\frac{No.OfFavourableOutcomes}{TotalNo.OfOutcomes}[/tex]

a) Divisible by 5 and not by 7 means, that the number is divisible by 5 but not by 35.

Total numbers between 1000 and 9999 divisible by 5= 1800

Total numbers between 1000 and 9999 divisible by 35=257

Total numbers between 1000 and 9999 divisible by 5 but not by 35=         1800 - 257= 1543

Probability=[tex]\frac{1543}{9000}[/tex]=0.1714

b) Finding the number of numbers with distinct digits,

We can find this by the method that how many digits can be placed in each place,

In the thousands place=9( 0 cannot be placed)

In the hundreds place=9( The digit inserted in the thousands place, cannot be put but 0 can be put now)

In the tens place=8( Numbers placed in hundreds and thousands place cannot be put)

In the ones place=7(The rest 3 numbers can't be put)

Total Favorable Outcomes=9 x 9 x 8 x 7 =4536

Probability=[tex]\frac{4536}{9000}[/tex]=[tex]\frac{63}{125}[/tex]=0.504

c) Total number of even numbers between 1000 and 9999= 4500

Probability=[tex]\frac{4500}{9000}[/tex]=[tex]\frac{1}{2}[/tex]=0.5

d) Total numbers between 1000 and 9999 divisible by 5= 1800

Probability= [tex]\frac{1800}{9000}[/tex]=[tex]\frac{1}{5}[/tex]=0.2

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