smc

The probability that Rachelle gets a C in English is $1-\frac{1}{6}-\frac{1}{4}=\frac{7}{12}$. The probability that she gets a C in History is $1-\frac{1}{4}-\frac{1}{3}=\frac{5}{12}$. We see that the sum of Rachelle's "point" scores must be at least 14 since $4\ast 3.5=14$. We know that in Mathematics and Science we have a total point score of 8 (since she will get As in both), so we only need a sum of 6 in English and History. This can be achieved by getting two As, one A and one B, one A and one C, or two Bs. We evaluate these cases. The probability that she gets two As is $\frac{1}{6}\cdot \frac{1}{4}=\frac{1}{24}$. The probability that she gets one A and one B is $\frac{1}{6}\cdot \frac{1}{3}+\frac{1}{4}\cdot \frac{1}{4}=\frac{1}{18}+\frac{1}{16}=\frac{8}{144}+\frac{9}{144}=\frac{17}{144}$. The probability that she gets one A and one C is $\frac{1}{6}\cdot \frac{5}{12}+\frac{1}{4}\cdot \frac{7}{12}=\frac{5}{72}+\frac{7}{48}=\frac{31}{144}$. The probability that she gets two Bs is $\frac{1}{4}\cdot \frac{1}{3}=\frac{1}{12}$. Adding these, we get $\frac{1}{24}+\frac{17}{144}+\frac{31}{144}+\frac{1}{12}=\frac{66}{144}=\boxed{\frac{11}{24}}$.