Total number of outcomes =‌52C2=1326 Let A={4,8,10},B={6,9} and C={3}A′= Cards which have the number either 4 or 8 or 10 B′= Cards which have the number either 6 or 9 C′= Cards which have the number 3 Now, ‌n(A′)=3×4=12 ‌n(B′)=2×4=8 ‌n(C′)=1×4=4 Favourable cases Case I 0 cards from A′,1 cards from B′ and 1 card from C′. Case II 0 cards from A′,2 cards from B′ and 0 cards from C′ Case III 1 cards from A′,1 card from B′ and 0 cards from C′ Case IV 1 cards from A′,0 cards from B′ and 1 card from C′
Total number of cases ‌=‌12C0×‌8C1×‌4C1+‌12C0×‌8C2×‌4C0 ‌+‌12C1×‌8C1×‌4C0+‌12C1×‌8C0×‌4C1 ‌=1×8×4+1×28×1+12×8×1+12×1×4 ‌=32+28+96+48=204 ‌∴‌ Required probability ‌=‌