A 4 digit lottery # is to be chosen from among the digits {1,2,3,4,5,7,8,9} where no digit may be used more?

Question

than once.

a. What is the probability that the lottery number contains only odd digits?
b. What is the probability that the lottery number begins with 1 and ends with 9?

Answer 1

assuming 6 is absent

ans a= 5P4 / 8P4 = .07142 (as v hav 5 odd nos and only 4 digit no is to b formed)

ans b = 6P2 / 8P4= .017857 ( 1 _ _ 9 . V hav only 6 nos left 2 fill 2 places)


assuming u missed 6

ans a= 5P4 / 9P4= .03968 (as v hav 5 odd nos and only 4 digit no is to b formed)

ans b = 7P2 /9P4 = .013888 ( 1 _ _ 9 . V hav only 7 nos left 2 fill 2 places)

Answer 2

I'm assuming that 0 and 6 are not options, given your problem:

FIRST PART:
The probability that your first digit is odd is 5/8 (there are 5 odd digits out of 8 digits in your initial set)
The probability that the second digit is odd is 4/7 (4 odd digits remaining out of 7 digits)
The probability that the third digit is odd is 3/6 (3 odds in 6 digits)
The probability that the fourth digit is odd is 2/5 (2 odds in 5 digits).

So the probability that the lottery number contains only odd digits is: 5/8 x 4/7 x 3/6 x 2/5

Note this is the same as C(5,4)/C(8,4). The number of ways to arrange 4 odd digits out of 5 / the total number of way to pick 4 digits out of 8.

This is 120/1680 which reduces to 1/14
a. 1/14

SECOND PART:
The probability that your first digit is 1 is 1/8 (1 choice in your initial set of 8)
The probability that the second digit is not 1 or 9 is 6/7 (6 digits are not 1 or 9, out of the 7 digits)
The probability that the third digit is not 1 or 9 is 5/6 (5 digits are not 1 or 9, out of the remaining 6 digits)
The probability that the fourth digit is 9 is 1/5 (1 choice in the remaining 5 digits).

So the probability that the lottery number starts
is: 1/8 x 6/7 x 5/6 x 1/5

This is also C(6,2) / C(8,4) or the ways to arrange the middle two digits in 1xx9 out of the two combinations of 8 balls

This is 30/1680 which reduces to 1/56
b. 1/56

So given your stated problem with digits 1, 2, 3, 4, 5, 7, 8, 9 the answers are:
a. 1/14
b. 1/56

ON THE OTHER HAND:
If you meant to include all the digits (0 through 9) then we have a whole different problem.

Following the pattern above you have:
5/10 x 4/9 x 3/8 x 2/7
C(5,4) = 5x4x3x2
C(10,4) = 10x9x8x7
a. 1/42

For the second part:
1/10 x 8/9 x 7/8 x 1/7
C(8,2) = 8x7
C(10,4) = 10x9x8x7
b. 1/90

FINAL SUMMARY:
If using only {1, 2, 3, 4, 5, 7, 8, 9}
a. 1/14
b. 1/56

If using all digits {0, ..., 9}
a. 1/42
b. 1/90

Answer 3

a) the probability of the odd numbers is about 20.
b) the probability of the numbers between 1 to 9 is about 36.

Answer 4

I agree with bandf's method, but his calculation on the part b is wrong. It's 1/56. 1/8 * 6/7 * 5/6 * 1/5 = 1/8 * 1/7 (all the 5s and 6s cancel out).

Answer 5

a) 1 in 14
b) 1 in 56

Answer 6

A. I am not exactly sure. I think 20% or something like that!
B.: 1579;1539;1359;1379;1759;1739
Am I right???