Candice returns home

After rattling her brain with combinatorics, Candice finally decides how she wants to buy her candy. She bought what she wanted and was on her way home when she stumbles across a $ 100 \$100 note on the pavement.

Candice, being the innocent little child she is, decides to take the note. She notices that the note is numbered A W K 45 AWK45 .

If the probability that she has touched this note before is x x , evaluate 1 0 5 x \lfloor 10^5 x \rfloor .

Details and assumptions

  • Each note is labelled in the form L L L N N LLLNN where L represents a letter and N represents non-negative integer less than 10.
  • Assume this currency is limited to only $ 100 \$100 notes and that each possible note is currently in circulation. There is equal probability of this being any one of the notes.
  • Candice is 10 years old. Assume that every note her mother earns from the moment she is born touches her hand. These do not have to be unique.
  • Her mother makes exactly $ 100000 \$100000 a month.
  • You are allowed to use a calculator. Do not make any extra assumptions.
Thanks to Calvin Lin for helping with the problem.
Image Credit : barryfromtexas on Photobucket .
Check out Candice's Other Adventures!


The answer is 6599.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Vishnu Bhagyanath
Aug 22, 2015

Candice is 10 years old, she has touched 1000 1000 bills each month for 12 months in 10 years. So in total she has touched 1000 12 10 = 120000 1000 * 12 * 10 = 120000 notes (not necessarily unique notes).

Since the city has a total of all possible combination of LLLNN notes, there can be 2 6 3 × 1 0 2 26^3 \times 10^2 notes.

The probability that this note has N O T NOT been touched by her would be ( 1 1 2 6 3 1 0 2 ) 120000 (1- \frac{1}{26^310^2})^{120000} .

Therefore, the probability that she H A S HAS touched that note is 1 ( 1 1 2 6 3 1 0 2 ) 120000 1- (1- \frac{1}{26^310^2})^{120000} .

Converting the answer to the given form, ( 1 ( 1 1 2 6 3 1 0 2 ) 120000 ) × 1 0 5 = 6599 \lfloor (1- (1- \frac{1}{26^310^2})^{120000}) \times 10^5 \rfloor= 6599

Something seems wrong here. Probability of her 'not' touching that note greater than 1? And probability of her touching that note is negative?

Siva Bathula - 5 years, 2 months ago

I got 6827...

Finn C - 2 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...