Suppose that a 4-digit pin number is to be formed with the digits in {0, 1, ,2 ,3 ,4 ,5, 6, 7, 8, 9}. What is the probability that the pin number contains at le

Question

Suppose that a 4-digit pin number is to be formed with the digits in {0, 1, ,2 ,3 ,4 ,5, 6, 7, 8, 9}. What is the probability that the pin number contains at least 3 odd digits?

Answer 1

Since you are looking for a PIN number order counts (permutations).

 

At least 3 odd numbers means you want 3 or all 4 to be odd.

 

P(at least 3) = P(exactly 3) + P(all 4)

 

 

three odd can be OOOE  or  OOEO  or OEOO or EOOO  so this is 4(5*5*5*5) ways

 

all 4 odd would be OOOO = 5*5*5*5 ways

 

so the total number of ways is the sum of those which is 5(5*5*5*5)

 

That would be the numerator and 10000 would be the denominator. 

 

 

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Answer 2

The total number of possible combinations if found by multiplying the possible number of options for each digit. Since there are 10 possible numbers for each digit, the total number of possibilities would be:

 

10 * 10 * 10 * 10 = 10^4 = 10,000

 

To find the possible outcomes containing at least three odd number will be found in a similar manner. There are 10 possible numbers and 5 possible odd numbers. Since three of the digits need to be odd and one digit can be any number, the number of possibilities would be: 

 

5 * 5 * 5 * 10 = (5^3) * 10 = 1,250

 

The probability would be the ratio of pin numbers containing at least three odd digits to the total number of possible pin numbers:

 

1250 / 10000 = 125 / 1000