Given: A bag containing 7 white, 5 black and 4 red balls.

By using the formula,

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

Two balls are drawn at random, therefore

Total possible outcomes are ^{16}C_{2}

n (S) = 120

**(i)** Let E be the event of getting both white balls

E = {(W) (W)}

n (E) = ^{7}C_{2 }= 21

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

=\( \dfrac{21}{120}\)

= \( \dfrac{7}{40}\)

**(ii)** Let E be the event of getting one black and one red ball

E = {(B) (R)}

n (E) = ^{5}C_{1}^{4}C_{1 }= 20

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

= \( \dfrac{20}{120}\)

= \( \dfrac{1}{6}\)

**(iii)** Let E be the event of getting both balls of same colour

E = {(B) (B)} or {(W) (W)} or {(R) (R)}

n (E) = ^{7}C_{2}+^{5}C_{2}+^{4}C_{2 }= 37

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

= \( \dfrac{37}{120}\)

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