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Question -

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



Answer -

Given: A box contains100bulbs, 20 of which are defective.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

Ten bulbs are drawn atrandom for inspection,

Total possibleoutcomes are┬а100C10

n (S) =┬а100C10

(i)┬аLet E be theevent that all ten bulbs are defective

n (E) =┬а20C10

P (E) = n (E) / n (S)

=┬а20C10┬а/┬а100C10

(ii)┬аLet E be theevent that all ten good bulbs are selected

n (E) =┬а80C10

P (E) = n (E) / n (S)

=┬а80C10┬а/┬а100C10

(iii)┬аLet E be theevent 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 defectivebulbs

Let EтА▓ be the eventthat none of the bulb is defective

n (EтА▓) =┬а80C10

P (EтА▓) = n (EтА▓) / n(S)

=┬а80C10┬а/┬а100C10

So, P (E) = 1 тАУ P (EтА▓)

= 1 тАУ┬а80C10┬а/┬а100C10

(iv)┬аLet E be theevent that none of the selected bulb is defective

n (E) =┬а80C10

P (E) = n (E) / n (S)

=┬а80C10┬а/┬а100C10

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