All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Probing the prob. (Posted on 2014-01-21)
Chose at random two distinct decimal digits.
Denote them as A & B.

1. What is the probability that:AB is less than A+B?
2. Same question, but with the word "distinct" erased.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 3 of 5 |

Part 1

If the lower number is 0, all nine higher digits will satisfy the condition.

If the lower number is 1, all eight higher digits will satisfy the condition.

If the lower number is 2 or higher, none of the higher digits will satisfy the condition.

So the probability is (9+8)/C(10,2) = 17/45 ~= .3777777777777778.

Part 2

In part 1 every pair was equally likely, but with duplication allowed, pairs of equal numbers are each half as likely as a specific pair of unequal numbers.

Effectively the above numbers for unequal digits can be doubled (1 and 8 counts as does 8 and 1).

For equal digits: Only (1,1) satisfies the condition.

The probability is now 35/10^2 = 35/100 = 7/20 = .35.

 Posted by Charlie on 2014-01-21 12:51:18

 Search: Search body:
Forums (0)