Vivekanand Vellanki
1

Let A be the probability that there is exactly one head; and

Let B be the probability that the die rolled 1, 2, 3, or 4.

The question asks for P(B/A). Using Bayes theorem:

P\left(\frac{B}{A}\right)\ =\ \frac{P\left(\frac{A}{B}\right)\cdot P\left(B\right)}{P\left(A\right)}

P(A/B) = 1/2

P(B) = 4/6

P(A) = 2/6*3/8 + 4/6*1/2 = 1/8+1/3 = 11/24

P\left(\frac{B}{A}\right)=\frac{\left(\frac{1}{2}\cdot\frac{4}{6}\right)}{\frac{11}{24}}=\frac{1}{2}\cdot\frac{4}{6}\cdot\frac{24}{11}=\frac{8}{11}