Given: A box contains 100bulbs, 20 of which are defective.

By using the formula,

P (E) = \( \dfrac{favourable\; outcomes }{total\; possible\; outcomes}\)

Ten bulbs are drawn at random for inspection,

Total possible outcomes are ^{100}C_{10}

n (S) = ^{100}C_{10}

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

n (E) = ^{20}C_{10}

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{ ^{20}{C}_{10}}{^{100}C_{10}}\)

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

n (E) = ^{80}C_{10}

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

= \(\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)

**(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′) = ^{80}C_{10}

P (E′) = \( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)

So, P (E) = 1 – P (E′)

=\( 1-\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)

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

n (E) = ^{80}C_{10}

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)

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