What is the probability that a person has both of these attributes? The probability that a blue-eyed person is left-handed is $\frac{1}{7}$.
The probability that a left-handed person is blue-eyed is $\frac{1}{3}$ . 
The probability that a person has neither of the attributes is $\frac{4}{5}$. 
What is the probability that a person has both attributes?
I just don't know how they got the answer
Answer: $\frac{1}{45}$
 A: Hint:
$P\left[L\mid B\right]=\frac{1}{7}$ i.e. $P\left[L\cap B\right]=\frac{1}{7}P\left[B\right]$
$P\left[B\mid L\right]=\frac{1}{3}$ i.e. $P\left[L\cap B\right]=\frac{1}{3}P\left[L\right]$
$1-P\left[L\cup B\right]=\frac{4}{5}$ 
These equations are enough to find $P\left[L\cap B\right]$
A: Use $P(A | B) = P(A \cap B)/P(B)$. You are given $P(L|B) = \frac 17$ and $P(B|L) = \frac 13$. Thus $P(B) = 7 P(B \cap L)$ and $P(L) = 3 P(B \cap L)$.
You are also given $P(B \cup L) = \frac 15$ since this is the probability of having either attribute. Since $P(B \cup L) + P(B \cap L) = P(B) + P(L)$ you obtain $$P(B \cap L) + \frac 15 = 10 P(B \cap L).$$
A: Let $P(L)$ be the probability of a person is left-handed and $P(B)$ the probability that a person is blue-eyed.
Then we know:
$$\frac{1}{7} = P(L|B) = \frac{P(L\cap B)}{P(B)}$$
$$\frac{1}{3} = P(B|L) = \frac{P(L\cap B)}{P(L)}$$
$$\frac{4}{5} = 1 - P(L\cup B) = 1 + P(L \cap B) - P(L) - P(B)$$
So $$P(L) = 3P(L\cap B)$$
$$P(B) = 7P(L\cap B)$$
Therefore $$\frac{4}{5} = 1 + P(L \cap B) - 3P(L \cap B) - 7P(L\cap B) = 1 - 9P(L\cap B) \Rightarrow \\ \Rightarrow 9P(L\cap B) = \frac{1}{5} \Rightarrow P(L\cap B) = \frac{1}{5·9} = \frac{1}{45}$$
A: Draw a Venn diagram!
Let probabilities be as follows:
$a$=Blue only
$b$=Blue and Left
$c$=Left only
$d$=None
From probabilities given, 
$$\begin{align}
\frac b{a+b}=\frac 17 \quad \Rightarrow a&=6b\\
\frac b{b+c}=\frac 13 \quad \Rightarrow c&=2b\\
d&=\frac 45\\
\end{align}$$
As the probabilites sum to 1, 
$$\begin{align}
a+b+c+d&=1\\
6b+b+2b+\frac 45&=1\\
9b&=\frac 15\\
b&=\frac 1{45}\end{align}$$
