Question

Irving cannot remember the correct order of the five digits in his ID number. He does remember that the ID number contains the digits 1, 4, 3, 7, 6. What is the probability that the first three digits of Irving's ID number will all be odd numbers?

Answer 1

Answer: The probability is 0.10

Explanation:

The ID number has five digits, and the digits can be 1, 4, 3, 7 and 6, and I will assume that each digit appears only once.

Then if we want to calculate the probability that the first 3 digits will be odd is:

Suppose that we have 5 slots, we want that in the first two slots to have odd numbers.

In our set, we have only 3 odd numbers {1, 3 and 7}

Then if we want an odd number in the first digit, we have 3 options

If we want an odd number in the second digit, we have two options (because we already selected one in the first selection)

If we want an odd number in the first digit, we have only one option.

For the fourth digit we have one of the two remaining even options, so we have 2 options.

For the fifth digit, we have only one digit.

The number of combinations is equal to the product of the number of options in each selection:

c = 3*2*1*2*1 = 12

Now, the total number of combinations is:

For the first digit we have 5 options

for the second digit we have 4 options.

for the third digit we have 3 options.

for the fourth digit we have 2 options.

for the fifth digit we have 1 options.

The number of combinations is:

C = 5*4*3*2*1 = 120.

Then the probability that the first 3 digits are odd numbers, is equal to the quotient between number of combination that start with 3 odd digits and the total number of combinations:

P = c/C = 12/120 = 0.10

Answer 2

Answer: 1/10

Explanation:

given:

numbers contained in the i.d

1,4,3,7,6

1. permutations of 5 possible outcome

T = 5

= 5 * 4 * 3 * 2 *1

= 120 times.

2. permutations of 3 odd numbers

( 1,3 and 7 )

T = 3

= 3 * 2 * 1 * 2 * 1

= 12

probability of of first three digits being odd numbers

P = 12 / 120

= 1 / 10