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Four pencils are chosen (without replacement) from a pencil case, two of which are blunt. The probability that both blunt pencils are chosen is twice the probability that neither is chosen. How many pencils are there in the pencil case?

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  • $\begingroup$ Why the close vote? Too simple? $\endgroup$
    – Simd
    Nov 30, 2023 at 10:33
  • $\begingroup$ Sorry, but this sort of basic maths problem is off-topic. $\endgroup$ Nov 30, 2023 at 10:33
  • $\begingroup$ @Randal'Thor I thought it was silly enough to be a puzzle :-/ I can't delete it now in any case. $\endgroup$
    – Simd
    Nov 30, 2023 at 10:34
  • $\begingroup$ The “color” aspects - pencils instead of balls from an urn, say - don’t really change what type of a puzzle something is, though. There’s no real clever insight needed here, just plugging the proper formulas together. $\endgroup$
    – Sneftel
    Nov 30, 2023 at 10:36

1 Answer 1

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There are

seven pencils.

Let $n$ be the number of sharp pencils, then

we know that $\binom n2=2\binom n4$. This is satisfied for $n=5$ where $\binom 52=10$ and $\binom n4=5$.

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  • $\begingroup$ Very nice and very fast! $\endgroup$
    – Simd
    Nov 30, 2023 at 10:33

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