A room has three lamps .From a collectionof 10 light bulbs of which 6 are no good,a person selects 3 at random and puts the in socket.What is the probability that he will have the light?
On your first draw, the probability of getting a good bulb is 4/10.
on your second draw, the probability of getting a good bulb is 3/9.
on your third draw, the probability of getting a good bulb is 2/8.
the probability that you will get a good bulb on all 3 draws is therefore equal to 4/10 * 3/9 * 2/8 which is equal to 24/720 which can be simplified to 1/30 which has a decimal equivalent of .033 rounded to 3 decimal places.
you can also look at it as the number of possible ways you can get a good bulb divided by the total possible ways you can get a bulb.
the equation for that is c(4,3) / c(10,3) which is equal to 4 / 120 which is equal to 1/30 again.
note that c(4,3) and 4c3 mean the same thing.
i mention this now because i used the ncx terminology down below when i used the c(n,x) terminology here.
c(n,x) is the combination formula of n! / (x! * (n-x)!)
ncx is also the combination formula of n! / (x! * (n-x)!)
when n = 10 and x = 3, this formula becomes 10! / (3! * 7!) which becomes 10*9*8*7! divided by 3! * 7! which becomes 10*9*8 / 3! which becomes 10*9*8 divided by 3*2*1 which becomes 10*3*4 which becomes 120.
similar machinations for when n = 4 and x = 3 results in 4.
you have 2 formula that both point to the same conclusion.
this indicates a fair chance that the solution is good unless you completely misunderstood what the problem is asking you to do.
this can be confirmed with a simple example that the formula is good.
assume 4 bulbs of which 3 are good and 1 is bad.
you want to draw 2 bulbs.
what is the probability that both bulbs will be good.
the formula says that the probability will be 3/4 * 2/3 = 6/12 = 1/2.