For the events $A$, $B$ and $C$ it is given that $P(A) = P(A \cap B) = \tfrac{1}{5}$ and $P(B) = 2P(A)$. It is also given that $P(B \cap C) = \tfrac{3}{10}.$

1. Establish that $A$ is a subset of $B$ and find $P(A' \cap B)$.

Since $P(A \cap B) = \frac{1}{5}$ and $P(A) = \frac{1}{5}$, we have $P(A \cap B) = P(A)$, which means that event $A$ is entirely contained within event $B$ ($A \subseteq B$).
$P(B) = 2P(A) = 2 \left(\frac{1}{5}\right) = \frac{2}{5}$
The probability of the region in $B$ but not in $A$ is:
$P(A' \cap B) = P(B) - P(A \cap B) = \frac{2}{5} - \frac{1}{5} = \frac{1}{5}$

2. Find the greatest possible value of $P(A \cap C)$.

The probability of $A \cap C$ cannot exceed the probability of $A$ itself.
$P(A \cap C) \le P(A) = \frac{1}{5}$
Also, since $A \subseteq B$, $A \cap C$ must be a subset of $B \cap C$, so:
$P(A \cap C) \le P(B \cap C) = \frac{3}{10}$
Since $\frac{1}{5} \le \frac{3}{10}$, the maximum possible overlap between $A$ and $C$ is bounded by $P(A)$.
Greatest possible value = $\frac{1}{5}$ (occurring if $A \subseteq C$).

3. Relate $P(A \cap C)$ to $P(B \cap C)$ to set up finding the minimum bound.

Since $A \subseteq B$, the intersection $B \cap C$ can be exactly partitioned into the part inside $A$ and the part outside $A$:
$P(B \cap C) = P(A \cap C) + P(A' \cap B \cap C)$
Substituting the given value $P(B \cap C) = \frac{3}{10}$:
$P(A \cap C) = \frac{3}{10} - P(A' \cap B \cap C)$
To find the least value of $P(A \cap C)$, we must maximize $P(A' \cap B \cap C)$.

4. Maximize $P(A' \cap B \cap C)$ to calculate the least possible value of $P(A \cap C)$.

The intersection $A' \cap B \cap C$ is a subset of $A' \cap B$, so its probability is bounded by $P(A' \cap B)$:
$P(A' \cap B \cap C) \le P(A' \cap B) = \frac{1}{5}$
Therefore, the least possible value of $P(A \cap C)$ is:
$P(A \cap C) \ge \frac{3}{10} - \frac{1}{5} = \frac{1}{10}$

1. Deduce the subset relationship between A, B, and C for the greatest probability.

Since $P(A \cap B) = P(A) = \frac{1}{5}$, event $A$ is a subset of event $B$ ($A \subseteq B$).
The greatest possible value for $P(A \cap C)$ is bounded by $P(A)$, so the maximum is $\frac{1}{5}$. This occurs when event $A$ is also a subset of event $C$ ($A \subseteq C$).
Thus, $A$ must be a subset of both $B$ and $C$, meaning $A \subseteq (B \cap C)$.

2. Describe the resulting Venn diagram.

To visually represent this, we draw events $B$ and $C$ as intersecting circles.
Since $A \subseteq (B \cap C)$, circle $A$ is drawn entirely within the overlapping region of $B$ and $C$.

1. Find the probability of the union of all three events using the complement rule.

$P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C') = 1 - \frac{1}{12} = \frac{11}{12}$

2. Recognise that $A$ is a subset of $B$ to simplify the union.

Since $P(A) = P(A \cap B) = \frac{1}{5}$, all outcomes in $A$ are also in $B$, so $A \subset B$.
Therefore, $A \cup B \cup C = B \cup C$, which means $P(B \cup C) = \frac{11}{12}$.

3. Calculate $P(C)$ using the addition rule for $B$ and $C$.

$P(B \cup C) = P(B) + P(C) - P(B \cap C)$
$\frac{11}{12} = \frac{2}{5} + P(C) - \frac{3}{10}$
$P(C) = \frac{11}{12} - \frac{1}{10} = \frac{49}{60}$

4. Calculate $P(A \cap C)$ using the definition of independent events.

Since $A$ and $C$ are independent events,
$P(A \cap C) = P(A)P(C)$
$P(A \cap C) = \frac{1}{5} \times \frac{49}{60} = \frac{49}{300}$