\(P(A\cap B')=P(A)-P(A\cap B)=0.54-0.35=0.19\)

Total number of persons  = 12, which can be arranged in a row in 12! ways.
Now, if 6 girls sit together, then we have 7 persons (Considering 6 girls as one person), which can be in a row in 7! ways. But 6 girls can be arrange in 6! ways. 
\(\therefore\)  Required probability
\(\frac { 7!\times 6! }{ 12! } =\frac { 1 }{ 132 } \)

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 =  \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 5040 } \)

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]
= \(\frac { 1 }{ 11 } +\frac { 2 }{ 11 } +\frac { 1 }{ 11 } =\frac { 4 }{ 11 } \)            

 Ans.Yes.