Given apples are: 3 bad, 15 good
Let X be no. of bad apples.
⇒P(X=0)=18C215C2=2!×16!18!2!×13!15!
⇒P(X=0)=18×1715×14=3×175×7
⇒P(X=0)=5135
⇒P(X=1)=18C23C1×15C1=9×173×15
⇒P(X=1)=175
⇒P(X=2)=18C23C2=9×173
⇒P(X=2)=1533
⇒E(X)=0×5135+1×175+2×1533
⇒E(X)=31
We know that, Var(X)=E(X2)−(E(X))2
⇒Var(X)=0×5135+1×175+4×511−(31)2
⇒Var(X)=5119−91
⇒Var(X)=15340