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Question Number 58050 by Aditya789 last updated on 17/Apr/19

	Answered by tanmay last updated on 17/Apr/19
	$D_1 ={1 2 3 4 5 6 } D_2 ={1 2 3 4 5 6} D_3 ={1 2 3 4 5 6} D_1 =1 (say)→6C_1 D_2 ={2 3 4 5 6}any one face →5_C_1 say D_2 =2 →4C_1 p(A)=((6×5×4)/(6×6×6))=(5/9) plz check is it?correct..$
	$D_{1} = {1 2 3 4 5 6}$ $D_{2} = {1 2 3 4 5 6}$ $D_{3} = {1 2 3 4 5 6}$ $D_{1} = 1 (s a y) \to 6 C_{1}$ $D_{2} = {2 3 4 5 6} a n y o n e f a c e \to 5_{C_{1}}$ $s a y D_{2} = 2 \to 4 C_{1}$ $p (A) = \frac{6 \times 5 \times 4}{6 \times 6 \times 6} = \frac{5}{9}$ $p l z c h e c k i s i t ? c o r r e c t . .$
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