3 Students, $A, B, C$ answer a question (independently). The probability of $A$ getting the correct answer is $0.9$, probability for $B$ is $0.7$ and the probability for $C$ is $0.4$ Q: Find the probability exactly two students get the question correct.
We can have $AB, BC, AC$ as those who get it correct. The question is, what about $ABC$?
I have
$P(\text{Exactly two get it right}) = P(AB \cup BC \cup AC) = P(AB) + P(AC) + P(BC) – P(AB \cap AC) – P(AB \cap BC) – P(BC \cap AC) + P(AB \cap AC \cap BC) $
But isn't that just
$$= P(AB) + P(AC) + P(BC) – 2 \cdot P(AB \cap AC \cap BC) $$
$$= (0.9*0.7) + (0.9*0.4) + (0.7*0.4) – 2(???)$$
Can we assume $AB, AC, BC$ are independent? What do I do?
Best Answer
P(A right, B right, C wrong): $0.9 \times 0.7 \times 0.6=0.378$
P(A right, B wrong, C right): $0.9 \times 0.3 \times 0.4=0.108$
P(A wrong, B right, C right): $0.1 \times 0.7\times 0.4=0.028$
Required probability: $0.378+0.108+0.028=0.514$