Answer:
0.7671 = 76.71% probability that he was taught by method A
Explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = (P(A \cap B))/(P(A))](https://img.qammunity.org/2022/formulas/mathematics/college/r4cfjc1pmnpwakr53eetfntfu2cgzen9tt.png)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Person learned Spanish successfully.
Event B: Method A was used.
Probability of a person learning Spanish successfully:
70% of 80%(using method A)
85% of 20%(using method B)
So
![P(A) = 0.7*0.8 + 0.85*0.2 = 0.73](https://img.qammunity.org/2022/formulas/mathematics/college/dnnmqka07wom3qh80ualwldkm11f31fslp.png)
Probability of a person learning Spanish successfully and using method A:
70% of 80%, so:
![P(A \cap B) = 0.7*0.8 = 0.56](https://img.qammunity.org/2022/formulas/mathematics/college/dlbbkt16pbdrluowosy95yaz1mgol1fkfx.png)
What is the probability that he was taught by method A?
![P(B|A) = (P(A \cap B))/(P(A)) = (0.56)/(0.73) = 0.7671](https://img.qammunity.org/2022/formulas/mathematics/college/324ikvsb5n86qol6mse2nhm411ohtp0eid.png)
0.7671 = 76.71% probability that he was taught by method A