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This question shows up in the binomial distribution chapter in my book.

g. How many men would you have to survey in order to be at least 95% sure you would find at least $1$ who is colorblind?

Now I know that $P(X \geq 1) = 1 - P(X=0) = 1 - {8\choose0}.0833^0.9167^8 = 1 - .4986752922 = .5013247078$

But this is for a sample of 8 men. This question is asking me to find $n$ with $95$% confidence,

But this seems like a confidence interval question which I'm a little confused about since my book is about probability and not about statistics. Nothing like this shows up in my book, so I'm stuck. I feel like this is a "challenge" question.

Can someone give me a clue on what formula to use to deal with a question like this? Is this even a confidence interval question?

Thank You

  • The question doesn't make sense because it starts with "you survey $8$ men..." and then asks "how many men would you have to survey...". If this were a typical confidence interval question, it wouldn't have the sentence with the $8$. – angryavian Jan 09 '21 at 19:44
  • @angryavian This could be one of the parts in a multi-part question. The actual question begins with "g.". Perhaps the OP forgot to filter out the redundant info. – Shubham Johri Jan 09 '21 at 19:45

1 Answers1

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You would have to solve $1 - {n\choose0}(0.0833^0)(0.9167^n)\ge0.95$.

Shubham Johri
  • 17,659
  • Forgive me, but when I plug in $86$ into the above, I get the following $1-({86 \choose 0} \cdot 0.0833^0 \cdot 0.9167^{86}) = .9994$ (approximately). I found that $n = 35$, is the minimum as $1-({35 \choose 0} \cdot 0.0833^0 \cdot 0.9167^{35}) = .9523$ (approximately). Where have I gone wrong? –  Jan 09 '21 at 20:00
  • Oh I am so sorry. I solved $1-\binom n10.9167^n\ge0.95$. That is the wrong inequality altogether. $n\ge35$ is absolutely correct. – Shubham Johri Jan 09 '21 at 20:01
  • ah okay, thank you for your help! –  Jan 09 '21 at 20:02