Answer :

given: box containing 6 red marbles numbered 1-6, 4 white marbles numbered 12-15

formula:


one marble is drawn from the given box, total possible outcomes are 10C1


therefore n(S)=10C1=10


(i) let E be the event of getting white marble


n(E)= 4C1=4




(ii) let E be the event of getting white marble with odd numbered


E= {13,15}


n(E)= 2




(iii) let E be the event of getting even numbered marble


E= {2, 4, 6, 12, 24}


n(E)= 5




(iv) let E1 be the event of getting red marble


(from (i))


Let E2 be the event of getting even numbered marble


(from (ii))


Therefore (E1 E2) = red coloured and even numbered


n (E1 E2) =3



By law of addition P(E1 E2) = P(E1)+P(E2)- P(E1 E2)



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