My question: Consider the numbers $1,2,3 ...,n$. All combinations of five of these numbers are written, and one combination is chosen at random. If the probability that this chosen combination does not contain the number $7$ is $0.875$, determine the value of $n$.

My problem: What should I do and how should I lay this problem out (all I can think of is a simple tree diagram?)? How do I create an equation where I can make $n$ the subject?

The most I can do: Is it saying there are $n! ,n! ,n! ,n! ,n!$? So choosing $1$ out of $5$ would be $1/5$. In $1/5$, I have the probability of not containing $7$ $(7/8)$ and $7$ $(1/8)$.

  • 2
    $\begingroup$ What does your concept of combination mean? Does it take order into account, allow repetition? $\endgroup$ – Mindlack Feb 20 at 12:42
  • $\begingroup$ @Mindlack exactly my dood, i knew someone going to ask, but i copied down exactly how the question on my textbook is written $\endgroup$ – Fred Weasley Feb 20 at 12:48
  • $\begingroup$ Isn’t there a definition in the book someplace before? $\endgroup$ – Mindlack Feb 20 at 12:51

I assume that a combination is a set of 5 distinct elements whose order does not matter. In other words, sets $\{1,2,3,4,5\}$ and $\{5,4,3,2,1\}$ represent the same combination.

The total number of combinations is $\binom{n}{5}$. The number of combinations with one particular number missing (say 7) is $\binom{n-1}{5}$.

The probability that a number is missing in a randomly selected combination is therefore:

$$\frac{\binom{n-1}5}{\binom{n}5}=\frac{\frac{(n-1)(n-2)(n-3)(n-4)(n-5)}{5!}}{\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}}=\frac{n-5}n=\frac78\implies n=40$$


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