# Thread: 2 probability problems

1. ## 2 probability problems

a) 16 men were at work demolishing an old factory when part of it collapsed, injuring 3 of them. Three of the men were apprenticies

Calculate the probability that all 3 men were apprenticies??

b) Out of 40 paving slabs, 12 have internal cracks.

Determine the probability that, in choosing 5 slabs at random, a cracked slab is not selected….

answers for a is 1/560
b = 105/703

but how??

2. $\left( {\begin{array}{*{20}c}
{16} \\
3 \\

\end{array} } \right) = \frac{{16 \cdot 15 \cdot 14}}
{{3 \cdot 2 \cdot 1}} = 560$

There 560 ways to choose 3 men from 16.

3. ## :)

Originally Posted by Plato
$\left( {\begin{array}{*{20}c}
{16} \\
3 \\

\end{array} } \right) = \frac{{16 \cdot 15 \cdot 14}}
{{3 \cdot 2 \cdot 1}} = 560$

There 560 ways to choose 3 men from 16.
ah I see its a dependant event

hmm going to try second one again