the answer to the 4th question is..
total no. of ways 4^6
then the fav. cases...
for at least 5 =prob of 5 people + prob of 6 people
prob. of 5 people is no. of ways of choosin 5 people=6C5
then there will be three cases when 5 have same no. and 6th has the other no. thus for 4 nos. its 4*3
6C5 ( 4 * 3 ) ..
prob for 6 is only 4 cases..
hence reqd prob. is..
6C5*12 + 4 / 4^6
Moderation Team
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