Probability

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 ‌=‌

204

1326

=‌

102

663