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I Have $98$ pieces of metal.

$89$ are highly resistant and $9$ are poorly resistant.

Of the $89$ that are highly resistant, $73$ are highly conductive and $16$ are poorly conductive.

Of the $9$ that are poorly resistant, $5$ are highly conductive and $4$ are poorly conductive.

I need to determine the probability of all pieces being highly resistant if I choose $5$ in a row, without putting them back in after selection. I'm not quite sure where to start. I realize it's a combination problem with the probability of pieces that are resistant. Would doing ($89$ choose $5$) % ($98$ choose $5$) times $89/98$ be the correct way of going about it?

My idea is to grab the total possible combinations of highly resistant pieces and divided by the number of combinations possible and multiplying that bye the percentage of choosing a highly resistant piece?

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    $\dfrac{89}{98}$ is the probability that the first choice is highly resistant: it is in your $\dfrac{89\choose 5}{98 \choose 5}$ so you do not need it again – 2017-01-30

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My idea is to grab the total possible combinations of highly resistant pieces and divided by the number of combinations possible and multiplying that bye the percentage of choosing a highly resistant piece?

Good idea, but there are no percentages involved.

You require the probability of selecting exactly $5$ from the $89$ highly resistant pieces, when selecting any $5$ from all $98$ pieces.   All other information given in the problem is a red herring.

$\binom n r$ counts ways to select $r$ from $n$ items.   Put it together.