Question
Every student in a class speaks at least one of three languages. For every language, the probability that a random student speaks that particular language is $\frac{3}{4}$ and for every pair of languages, the probability that a random student speaks that pair is $\frac{1}{2}$. What is the probability that a student speaks all three languages?
My Attempt At a Solution:
Let $A$ be the event that a student speaks one language.
Let $B$ be the event that a student speaks two languages.
Let $C$ be the event that a student speaks three languages.
Then $A \subseteq B \subseteq C$.
We can find the number of students who only speak one language:
Many Thanks to user @N. F. Taussig
We see now that by in the inclusion-exclusion principle
$$\mathbb{P}(A\cup B \cup C)=\mathbb{P}(A)+\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(A \cap B)-\mathbb{P}(A \cap C)-\mathbb{P}(B \cap C)+\mathbb{P}(A \cap B \cap C) $$
$$1=3*(\frac{3}{4}-\frac{1}{4})+\mathbb{P}(A \cap B \cap C) $$ $$\implies \mathbb{P}(A\cap B \cap C)=\frac{1}{4}.$$
Thank you for your time.
Let's define our events differently.
Let the languages be $L_1$, $L_2$ and $L_3$. Let $\Pr(L_k)$ be the probability that a student speaks language $k$, $1 \leq k \leq 3$. We are given the following information:
By the Inclusion-Exclusion Principle, $$\Pr(L_1 \cup L_2 \cup L_3) = \Pr(L_1) + \Pr(L_2) + \Pr(L_3) - \Pr(L_1 \cap L_2) - \Pr(L_1 \cap L_3) - \Pr(L_2 \cap L_3) + \Pr(L_1 \cap L_2 \cap L_3)$$ Substitute the values stated above for $\Pr(L_1 \cup L_2 \cup L_3)$, $Pr(L_1)$, $\Pr(L_2)$, $\Pr(L_3)$, $\Pr(L_1 \cap L_2)$, $\Pr(L_1 \cap L_3)$, and $\Pr(L_2 \cap L_3)$ to determine $\Pr(L_1 \cap L_2 \cap L_3)$, the probability that a student speaks all three languages.