Question:
A box contains 10 transistors, of which 3 are defective. A transistor
is selected from the box and tested until a non-defective one is
chosen. If X is the number of transistors to be chosen. Find P(X = 3).
My attempt:
Since we keep testing until we find a non-defective one, then we need to select at least 1 non-defective transistor in our 3 tests.
then $P(X = 3) = $ $${7 \choose 1}{3 \choose 2} + {7 \choose 2}{3 \choose 1} \over {10 \choose 3}$$ = 7/10
However, final answer = 7/120, which contradicts my answer, and I'm quite confused.
Thanks in advance.
Best Answer
You stop when you pick a non-defective transistor. So for $X = 3$, the first two must be defective one's.
So the probability is,
$P(X = 3) = \displaystyle \frac{3}{10} \cdot \frac{2}{9} \cdot \frac{7}{8} = \frac{7}{120}$
Or the way you calculated, it should be
$ \displaystyle \frac {2! \cdot {3 \choose 2}{7 \choose 1} }{ 3! \cdot {10 \choose 3}} = \frac{7}{120}$
You choose two defective one's that can be arranged in the first and second place. The third place is fixed for the non-defective transistor. The denominator is arrangements of any three chosen transistors.