A box contains 100bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:

(i) all 10 are defective


(ii) all 10 are good


(iii) at least one is defective


(iv) none is defective

given: box with 100 bulbs of which, 20 are defective

formula:


ten bulbs are drawn at random for inspection, therefore


total possible outcomes are 100C10


therefore n(S)= 100C10


(i) let E be the event that all ten bulbs are defective


n(E)= 20C10




(ii) let E be the event that all ten good bulbs are selected


n(E)= 80C10




(iii) let E be the event that at least one bulb is defective


E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs


Let E’ be the event that none of the bulb is defective


n(E’) = 80C10




Therefore,


P(E)=1-P(E’)




(iv) let E be the event that none of the selected bulb is defective


n(E)= 80C10




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