I'm having difficulty understanding this problem and I was wondering if someone had an answer.

Let

[tex] n = \text{Number of dice} = 3[/tex]

[tex] p_s = \text{Single die success probability (when the face is 3 or more)} = \frac{1}{2}[/tex]

[tex] k = \text{Number of successful die} \in 0, 1, 2, 3[/tex]

[tex] P(n, k, p) = (^n_k) p^k(1-p)^{n-k}[/tex]

The probability of getting 2 success [tex] (k=2)[/tex] if each die is rolled only once [tex] (p = p_s = \frac{1}{2})[/tex] is

[tex] P(3,\ 2,\ \frac{1}{2}) = (^3_2)\left(\frac{1}{2}\right)^2\left(1-\frac{1}{2}\right)^{3-2} = \frac{3}{8} = 0.375[/tex]

If we have the option to reroll any failed die once,

then [tex] p = p_s + (1-p_s)p_s = \frac{3}{4}[/tex]

and [tex] P(3,\ 2,\ \frac{3}{4}) = (^3_2)\left(\frac{3}{4}\right)^2\left(1-\frac{3}{4}\right)^{3-2} = \frac{27}{64} = 0.421875[/tex]

What is the probability if we can reroll only

**one**failed die (instead of all 3)?