Hi, I need help with the following questions:

1) How many 10-digit numbers are there, such that the sum of the digits is divisible by 2?

2) We define a bow-tie quadrilateral as a quadrilateral where two sides cross each other. An example of a bow-tie quadrilateral is shown below.

Seven distinct points are chosen on a circle. We draw all \(\binom{7}{2} = 21\) chords that connect two of these points. Four of these 21 chords are selected at random. What is the probability that these four chosen chords form a bow-tie quadrilateral?

3) I have a hat that contains six nickels and five pennies. I draw coins from the hat at random, without replacement. Find the probability that after drawing four coins, I have removed at most two pennies from the hat.

4) The Boomtown Bears are playing against the Tipton Toros in a baseball tournament. The winner of the tournament is the first team that wins three games. The Bears have a probability of \(\frac{2}{3}\) of winning each game. Find the probability that the Bears win the tournament.

Thank you so much for your help!

Guest Aug 9, 2020

#1**0 **

1 - The first 10-digit number whose digit sum is divisible by 2 =1,000,000,001

The last 10-digit number whose digit sum is divisible by 2 =9,999,999,999

9,999,999,999 - 1,000,000,001 =8,999,999,998 / 2 + 1 = 4,500,000,000 - Total number of 10-digit numbers.

Guest Aug 9, 2020

#2**0 **

@answer #1 -- Thanks for your answer! However, I'm not sure I understand what you did!

I get that the smallest and the largest numbers that meet are the ones you listed. However, why are you subtracting them, then dividing that by 2 and adding 1? I'm confused, can you explain?

FYI for future answerers, I figured out #3, and #4, I don't need help and those. I still need help on #1 and #2. Thanks!

Guest Aug 9, 2020

#3**0 **

1 - How many terms divisible by 2 are there between 2 and 8??

You subtract the last from the first: 8 - 2 =6. This is the number of terms.

6 / 2 = 3. But 3 doesn't count both 8 and 2. When you are subtracting from each other, only one of them is

counted, so you must add 1 to count both of 8 and 2. So that you have: 2, 4, 6, 8 for a total of 4.

Exactly the same thing with 1,000,000,001 as the first term and 9,9999,999,999, the last term.

9,999,999,999 - 1,000,000,001 =8,999,999,998 /2 =4,499,999,999 + 1 =4,500,000,000.

Got it?

Guest Aug 9, 2020