1. **State the problem:** We need to find the probability that when one student is chosen at random from Group A and one from Group B, one student is above 160 cm and the other is below 155 cm. 2. **Identify the groups and their heights:** - Group A: 141, 148, 149, 151, 152, 153, 154, 161, 163, 170 - Group B: 142, 153, 153, 157, 158, 160, 162, 164, 178 (using x=3 from part a) 3. **Count the number of students in each group:** - Group A has 10 students. - Group B has 9 students. 4. **Define the events:** - Event A1: Student from Group A is above 160 cm. - Event A2: Student from Group A is below 155 cm. - Event B1: Student from Group B is above 160 cm. - Event B2: Student from Group B is below 155 cm. 5. **Count students satisfying each event:** - Group A above 160 cm: 161, 163, 170 → 3 students - Group A below 155 cm: 141, 148, 149, 151, 152, 153, 154 → 7 students - Group B above 160 cm: 162, 164, 178 → 3 students - Group B below 155 cm: 142, 153, 153 → 3 students 6. **Calculate probabilities for each event:** - $P(A1) = \frac{3}{10}$ - $P(A2) = \frac{7}{10}$ - $P(B1) = \frac{3}{9} = \frac{1}{3}$ - $P(B2) = \frac{3}{9} = \frac{1}{3}$ 7. **Calculate the probability that one student is above 160 cm and the other is below 155 cm:** This can happen in two ways: - Student from Group A is above 160 cm and student from Group B is below 155 cm. - Student from Group A is below 155 cm and student from Group B is above 160 cm. So, $$ P = P(A1) \times P(B2) + P(A2) \times P(B1) $$ 8. **Substitute values:** $$ P = \frac{3}{10} \times \frac{1}{3} + \frac{7}{10} \times \frac{1}{3} = \frac{3}{30} + \frac{7}{30} = \frac{10}{30} = \frac{1}{3} $$ 9. **Final answer:** The probability that one student is above 160 cm and the other is below 155 cm is $\boxed{\frac{1}{3}}$.