# Probability Problem: A box contains 100 balls

A box contains 100 balls, numbered from 1 to 100. If two balls are selected at random and at the same time from the box, what is the probability that the numbers on the balls will be:

a) consecutives

b) 2 multiples of 6

c) odd and even

d) 2 divisors of 60

a) $P=\left(\dfrac{1}{100} \times \dfrac{1}{99}\right)100=\dfrac{1}{99}$

b) $P=\left(\dfrac{16}{100} \times \dfrac{15}{99}\right)= \dfrac{4}{165}$

c) $P=\left(\dfrac{50}{100} \times \dfrac{50}{99}\right)2=\dfrac{50}{99}$

d) $P=\left(\dfrac{10}{100} \times \dfrac{9}{99}\right)=\dfrac{1}{110}$

• For (a), what are the possible ways to select consecutive numbers? How many ways are there? – rogerl Oct 21 '17 at 23:33
• For (d), note that divisors of $60$ are $1,2,3,4,5,6,10,12,15,20,30,60$, and there are $12$ of them. – Mark Oct 21 '17 at 23:39
• For (a) you might consider drawing two balls from $\{1,2,3\}$ as a simpler example. It obviously has a probability of $\frac23$ of the pair being consecutive – Henry Oct 21 '17 at 23:47
• You are right Mark and @Henry with your example now I understand it. Thank you – B. David Oct 22 '17 at 9:21

A box contains $100$ balls, numbered from $1$ to $100$. If two balls are selected at random, what is the probability that the numbers on the balls will be consecutive?

There are $\binom{100}{2}$ ways to select two of the $100$ balls.

A pair of consecutive numbers is determined by the smaller of the numbers. There are $99$ pairs of consecutive numbers since the smaller number can be at most $99$. Hence, the desired probability is $$\frac{99}{\dbinom{100}{2}}$$

What is the probability that two multiples of $6$ are selected?

There are $16$ multiples of six less than or equal to $100$. Hence, the number of favorable cases is $\binom{16}{2}$. Therefore, the desired probability is $$\frac{\dbinom{16}{2}}{\dbinom{100}{2}}$$ Your answer is correct.

What is the probability that an even number and an odd number are selected?

We must select one of the $50$ even numbers and one of the $50$ odd numbers, so the desired probability is $$\frac{\dbinom{50}{1}\dbinom{50}{1}}{\dbinom{100}{2}}$$ Your answer for this question is also correct.

What is the probability that $2$ divisors of $60$ are selected?

As @Mark pointed out in the comments, there are $12$ divisors of $60$. They are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. Hence, the desired probability is $$\frac{\dbinom{12}{2}}{\dbinom{100}{2}}$$