Question #136970
A box of 10 light bulbs, 2 of which are faulty. A bulb is selected at random and put into a socket. If it works it is left there, if it is faulty it is discarded and a new bulb is selected. The process I continued until a bulb which works is found. Determine the probability distribution of X, the number of bulbs tried.
1
Expert's answer
2020-10-06T17:51:25-0400

Let X=X= the number of bulbs tried.


P(X=1)=0.8=45P(X=1)=0.8=\dfrac{4}{5}


P(X=2)=0.289=845P(X=2)=0.2\cdot \dfrac{8}{9}=\dfrac{8}{45}


P(X=3)=0.2191=145P(X=3)=0.2\cdot \dfrac{1}{9}\cdot1=\dfrac{1}{45}


P(X=4)=P(X=5)=P(X=6)=P(X=7)=P(X=4)=P(X=5)=P(X=6)=P(X=7)=

=P(X=8)=P(X=9)=P(X=10)=0=P(X=8)=P(X=9)=P(X=10)=0

x12345678910p(x)4/58/451/450000000\begin{matrix} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ p(x) & 4/5 & 8/45 & 1/45 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}


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