0
$\begingroup$

The question I am working through states:

Three digits are selected at random from the digits 1 through 10, without replacement. Find the probability that at least 2 digits are prime.

I feel as though I am close, but not quite hitting the mark with my solution:

My total number of selections would be $10C3$, and there are 4 prime numbers to pick from $\{2,3,5,7\}$.

Since I'm looking for at least 2 prime numbers, that means I am looking for the probability that I select 2, or 3 prime numbers.

So, I have \begin{align*} \textrm{P(2 prime numbers)} & \ or\ \textrm{P(3 prime numbers)}\\ \frac{4C2}{10C3} & + \frac{4C3}{10C3}\\ &= \frac{1}{12} \end{align*}

Should I instead be looking for the compliment? I feel as though it would be just as complex and not really save me any time [1-P(0 or 1 primes)].

Edit to add new solution (looks better!): \begin{align*} \textrm{P(2 prime numbers)} & \ or\ \textrm{P(3 prime numbers)}\\ \frac{4C2 \times 6C1}{10C3} & + \frac{4C3}{10C3}\\ &= \frac{1}{3} \end{align*}

  • 1
    Look again at $P(2~\text{prime numbers})$. How many selections of **three** numbers have two of them prime (*and one of them not prime*). – 2017-01-06
  • 0
    Okay, so I'm forgetting about the $6C1$...in this case since we're using and, it would imply multiply. So, $4C2 \times 6C1$ all divided by $10C3$? – 2017-01-06
  • 0
    Which would get me $\frac{1}{3}$...this seems right!? – 2017-01-06

2 Answers 2

3

P(exact $1$ prime) = $\dfrac{\dbinom{4}{1}\dbinom{6}{2}}{\dbinom{10}{3}}$
P(No prime) = $\dfrac{\dbinom{6}{3}}{\dbinom{10}{3}}$

Answer
$1-\dfrac{\dbinom{4}{1}\dbinom{6}{2}}{\dbinom{10}{3}}-\dfrac{\dbinom{6}{3}}{\dbinom{10}{3}}$

  • 0
    Your P(exactly 1 prime) is the probability of the first digit drawn being prime, you need to mutliply by 3 to allow for the different order you can draw the prime at. – 2017-01-06
  • 1
    @ladamas, it is not needed. if we take order into consideration, 3! will come at numerator and denominator and hence cancel. – 2017-01-06
1

You are correct for the chance of selecting three prime numbers, but for selecting exactly two you need to multiply by the number of ways to select the non-prime.