TITLE: lamps probability .
QUESTION [2 upvotes]: A room has $3$ lamp sockets. From a collection of $10$ bulbs $,6$ are fused .
A person selects $3$ at random and puts them in sockets. The probability that he will have the light is
$\bf{My\; Try::}$ Total no. of events $ = $ no. of selecting $3$ lamps out of total $10$ is $ \displaystyle \binom{10}{3}$
Now favourable events $ = $ selecting $3$ bulbs out of total $4$ blubs $\displaystyle = \binom{4}{3}$
So Required probability $\displaystyle  = \frac{4\cdot 3 \cdot 2}{10\cdot 9 \cdot 8} = \frac{1}{30}$
Is my solution is right, If not then how can we solve it,
Thanks

REPLY [0 votes]: If the favorable event reads "he takes at least one working bulb", then your answer isn't correct.
Indeed this probability reads $1-\frac{ 6}{ 10}\frac{ 5}{ 9}\frac{ 4}{ 8} = \frac 5 6$ which amount to compute $1-\mbox{Probability of taking only fused bulbs}$.
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