Given: $P(H)=0.25,\; P(+)=0.20,\; P(+\mid H')=0.05.$

Find sensitivity from total probability:

$$0.20 = P(+\mid H)\cdot0.25 + 0.05\cdot0.75 \implies P(+\mid H)=0.65.$$

So $P(-\mid H)=0.35,\; P(-\mid H')=0.95.$

Compute $P(-)$:

$$P(-)=0.35\cdot0.25+0.95\cdot0.75=0.80.$$

Apply Bayes for $P(H'\mid -)$:

$$P(H'\mid -)=\frac{P(-\mid H')P(H')}{P(-)}=\frac{0.95\cdot0.75}{0.80}=\frac{0.7125}{0.80}\approx0.890625\;(89.06\%).$$

Conclusion: the correct posterior ≈89.1%. None of the given choices match exactly; C (87%) is closest and D (93%) is incorrect.