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(i)
Required probability $=\frac{14! \times 2!}{15!} = \frac{2}{15}$

(ii)
Taking complement,
Required probability $= 1 - \frac{13!\times 3!}{15!} = \frac{34}{35}$

(iii)
Required probability $=\frac{12! \times 2! \times 3!}{15!} = \frac{2}{455}$

(iv)
Required probability $=\frac{12}{15} + \frac{1}{35} - \frac{2}{455} = \frac{43}{273}$

(v)
Required probability $=\frac{13! \times 2!}{14!} = \frac{1}{7}$

### KS Comments:

Feel free to comment if you have questions for this. There are several alternative methods to solving this.