Let's solve the problem step by step.

Part (a)

We are given the following probabilities:

• $P(E1) = P(E2) = 0.15$
• $P(E3) = 0.4$
• $P(E4) = 2P(E5)$

The sum of all probabilities in a sample space must equal 1:

$P(E1) + P(E2) + P(E3) + P(E4) + P(E5) = 1$

Substituting the given values:

$0.15 + 0.15 + 0.4 + P(E4) + P(E5) = 1$

This simplifies to:

$0.7 + P(E4) + P(E5) = 1$

Subtracting 0.7 from both sides:

$P(E4) + P(E5) = 0.3$

We also know that $P(E4) = 2P(E5)$. Let's substitute $P(E4) = 2P(E5)$ into the equation:

$2P(E5) + P(E5) = 0.3$

This simplifies to:

$3P(E5) = 0.3$

Solving for $P(E5)$:

$P(E5) = \frac{0.3}{3} = 0.1$

Now, substituting $P(E5)$ back into the equation for $P(E4)$:

$P(E4) = 2 \times 0.1 = 0.2$

So, the probabilities are:

• $P(E4) = 0.2$
• $P(E5) = 0.1$

Part (b)

We are given the following probabilities:

• $P(E1) = 0.3$
• $P(E1) = 3P(E2)$

Let $P(E2) = x$, then $P(E1) = 3x$. Since $P(E1) = 0.3$, we can solve for $x$:

$3x = 0.3$

$x = \frac{0.3}{3} = 0.1$

So, $P(E2) = 0.1$.

We know that the remaining simple events (E3, E4, E5) are equally probable, and their probabilities must add up to $1 - (P(E1) + P(E2))$:

$P(E3) + P(E4) + P(E5) = 1 - 0.4 = 0.6$

Since they are equally probable:

$P(E3) = P(E4) = P(E5) = \frac{0.6}{3} = 0.2$

Thus, the probabilities are:

• $P(E1) = 0.3$
• $P(E2) = 0.1$
• $P(E3) = 0.2$
• $P(E4) = 0.2$
• $P(E5) = 0.2$

Summary:

1. In part (a):

   • $P(E4) = 0.2$
   • $P(E5) = 0.1$

2. In part (b):

   • $P(E1) = 0.3$
   • $P(E2) = 0.1$
   • $P(E3) = 0.2$
   • $P(E4) = 0.2$
   • $P(E5) = 0.2$

Do you want further details or have any questions?

Here are 5 related questions:

1. How do you calculate the probability of an event not occurring?
2. What is the probability of at least one of the events $E3$, $E4$, or $E5$ occurring?
3. How would you find the expected value if each event had a numerical outcome?
4. What if $P(E4) = 3P(E5)$, how would that change your calculation in part (a)?
5. How do you determine if events are mutually exclusive?

Tip: The sum of probabilities in any sample space must always equal 1. This principle is key when solving for unknown probabilities.