Answer:
Number of Questions =3
Probability of giving a correct answer
[tex]=\frac{1}{3}[/tex]
Probability of giving two correct answers
[tex]=\frac{2}{3}[/tex]
Probability of giving all correct answers
[tex]=\frac{3}{3}\\\\=1[/tex]
Probability that at least one of them is correct
[tex]=_{1}^{3}\textrm{C}\\\\=\frac{3!}{(3-1)! \times1!}\\\\=3 \text{ways}\\\\=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\\\=\frac{1}{27}[/tex]
Probability that two of them is correct
[tex]=_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times2!}\\\\=3 \text{ways}\\\\=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\\\=\frac{8}{27}[/tex]
Probability that all of them is correct
[tex]=_{3}^{3}\textrm{C}\\\\=\frac{3!}{(3-3)! \times3!}\\\\=1 \text{way}\\\\=1[/tex]
So, Required probability
[tex]=\frac{1}{27} \times \frac{8}{27} \times 1\\\\=\frac{8}{729}[/tex]