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) $P=\left(\dfrac{10}{100} \times \dfrac{9}{99}\right)=\dfrac{1}{110}$
Is correct my answer?
$\endgroup$ 41 Answer
$\begingroup$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 $$\frac{\dbinom{12}{2}}{\dbinom{100}{2}}$$
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