A sample space consists of five simple events, E1, E2, E3, E4, and E5. 
a If P(E1) = P(E2) = 0.15, P(E3) = 0.4, and P(E4) = 2P(E5), find the probabilities of E4 and E5.
 b If P(E1) = 3P(E2) = 0.3, find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable

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.

A sample space consists of five simple events, E1, E2, E3, E4, and E5. a) If P(E1) = P(E2) = 0.15, P(E3) = 0.4, and P(E4) = 2P(E5), find the probabilities of E4 and E5. b) If P(E1) = 3P(E2) = 0.3, find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable.

Solve a probability problem involving a sample space with five simple events: E1, E2, E3, E4, and E5. This detailed solution explains how to find the probabilities of E4 and E5, given specific conditions, and how to calculate the probabilities when events are equally likely. Suitable for Grades 10-12.

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