Arun
Last Activity: 7 Years ago
Total no . of outcomes = 6 4
Favourable cases are:-when 4D and only 1 of 1D2D3D shows same number or only 2 of 1D2D3D shows same numberor all 3 of 1D2D3D shows same number.
No. of favorable outcomes=61C×31C×5×5+61C×32C×5+61C×33C
=61C+C335×32C+5×5×31C
So, Required probability=outcomes of .no Totaloutcomes favorable of .No=4633C+5×32C+5×5×31C61C
=6×2161+5×3+25×3×6=21691
