It is a two stage experiment. When a die is rolled, then it may show up any one of the six number 1, 2, 3, 4, 5, 6.

When the die shows up an odd number a coin is toasted twice and hence, corresponding to each of 1, 3 and 5, there are four possibilities.
When the die shows up an even number, a coin is tossed once and hence, corresponding to each of 2,4 and 6, there are only two possibilities.
Hence, the sample space is
S={2H,2T,4H,4T,6H,6T,1HH,1HT,1TH,1TT,3HH,3HT3TH,3TT,5HH,5HT,5TH,5TT}

P( not A  and not B) =  \(P(\overline { A } \cap \overline { B } )\)
\(=P(\overline { A\cup B } )=1-P(A\cup B)\)
\(=1-[P(A)+P(B)-P(A\cap B)]\)
\(=1-\left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } \right] =1-\left[ \frac { 2+4-1 }{ 8 } \right] \)
\(=1-\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)

 Ans.Yes.

There are 11 letters in the word PROBABILITY.
Since, a letter is selected randomly, therefore the possible outcomes are P, R, O, B, A, I, L, T, Y.
But here, outcomes are not equally likely
as P(letter P) = \(\frac { 1 }{ 11 } ;\) P (letter R) = \(\frac { 1 }{ 11 } \)
P(letter O)= \(\frac { 1 }{ 11 } ;\) P (letter B) = \(\frac { 2 }{ 11 } \)
[\(\because \) B repeated twice in the word]
P(letter A) = \(\frac { 1 }{ 11 } ;\)   P(letter I) = \(\frac { 2 }{ 11 } \)
P(letter L) = \(\frac { 1 }{ 11 } ;\)  P(letter T) = \(\frac { 1 }{ 11 } \)
and  P(letter Y) = \(\frac { 1 }{ 11 } \)
Now, let E be event of getting a vowel. Then outcomes favourable to E are A, I and O.
\(\Rightarrow \)  E={A,I,O}
Now,\(P(E)=P(A)+P(I)+P(O)\)
[by aximatixc approach to probability]
= \(\frac { 1 }{ 11 } +\frac { 2 }{ 11 } +\frac { 1 }{ 11 } =\frac { 4 }{ 11 } \)            

When the digits are not repeated, then first place may have one of 10 digits, the second 9, third 8 and fourth 7. Number of 4-digit numbers, n(S)=10 x 9 x 8 x 7 = 5040. Now lock can be opened only in 1 way.
\(\therefore\)  n(E) = 1
Hence, probability of opening the lock =  \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 5040 } \)