MHB What is the Probability That Adam Answers Correctly?

[anemone]

The probability that both Adam and Nicolas answer a question correctly is 18 percent. The probability that one, but not both, of them answers the question correctly is 54 percent. Given that Adam and Nicolas are operating independently, what is the probability that Adam answers the question correctly, given that Adam is more likely than Nicolas to answer correctly?

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[anemone]

Congratulations to lfdahl for his correct solution as shown below!:)

Spoiler

Define the events:

$A \equiv $ Adam answers correctly. Thus $A^c$ is the event, that Adam answers incorrectly.

$B \equiv$ Nicolas answers correctly. $B^c$: Nicolas answers incorrectly.

Adam and Nicolas are operating independently, i.e.:
\[(1). \;\;\; P(A \cap B)=P(A)\cdot P(B) = 0.18=p_1\]

The probability that one of them, but not both, answers correctly is:
\[(2). \;\;\; P(A^c \cap B) + P(A\cap B^c) = 0.54 = p_2\]

The above mentioned independency, implies that:

\[(3). \;\;\; P(A^c \cap B) = P(A^c)\cdot P(B)=(1-P(A))\cdot P(B)\\\\
(4). \;\;\; P(A \cap B^c) = P(A)\cdot P(B^c)= P(A)\cdot (1-P(B))\]

Inserting $(3)$ and $(4)$ into $(2)$ and combining $(2)$ and $(1)$ yields:

\[[P(A)]^2-(2p_1+p_2)\cdot P(A)+p_1 = 0 \;\; \; \Rightarrow \;\;\; P(A)\in \left \{ 0.60, 0.30 \right \}\]

Given, that Adam is more likely to answer correct, the probabilities are:
\[P(A)=0.60 \;\; \wedge \;\; P(B) = 0.30\]