minimum no of states and final states

Question

I solved both of these qs in a traditional way,by drawing nfa and then convert them to dfa..by tthis process ans shoulbe 4 and 2..but both of them are wrong..pls check..

Comments

I still do not know the formal way for this. DFA 1 is accepting all strings ending in 1- so just 2 states needed.

For DFA 2, it is asking only the no. of final states. So just 1 as it is accepting all strings containing "10".

Answer 1

I. $(0+1)^*1$, all strings ending with $1$, need only $2$ states 

II. $0+1)*10(0+1)^*$, all strings contain substring $10$, need $3$ states, and they are asking about final state, that is only 1. 

[Note:
1. all strings ending with $n$ length string or all string contain $n$ length substring, will always $n+1$ states in minimal DFA.
2. After conversion from NFA to DFA, there are maximum probability of minimization, always look after it.]

Comments

I did this mstake only,i didnt check the redundency

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Yes, actually, during nfa to dfa conversion, we get several new states ( from combinations of other states), there are full chances of redundant/equivalent states.

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Sir , How can we convert epsilon nfa into dfa ?? If any reference plz post here ...