A sample space consists of five simple events with P(E₁) = P(E₂) = 0.3, P(E3) = 0.1, and P(E4) = 2P(E5). Find the probabilities for simple events E4 and E5. Step 1 Recall that the sum of the probabilities for all simple events in a sample space equals 1. Since there are five simple events, E₁, E2, E3, E4, E5, this yields the follow P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1 We are given that P(E₁) = P(E₂) = 0.3 and P(E3) = 0.1. Substitute these values into the above equation and simplify. P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1 + P(E4) + P(E5) = 1 + P(E4) + P(E5) = 1 P(E4) + P(E5) = 0.3 +0.3 +
A sample space consists of five simple events with P(E₁) = P(E₂) = 0.3, P(E3) = 0.1, and P(E4) = 2P(E5). Find the probabilities for simple events E4 and E5. Step 1 Recall that the sum of the probabilities for all simple events in a sample space equals 1. Since there are five simple events, E₁, E2, E3, E4, E5, this yields the follow P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1 We are given that P(E₁) = P(E₂) = 0.3 and P(E3) = 0.1. Substitute these values into the above equation and simplify. P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1 + P(E4) + P(E5) = 1 + P(E4) + P(E5) = 1 P(E4) + P(E5) = 0.3 +0.3 +
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Chapter1: Starting With Matlab
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Question

Transcribed Image Text:Tutorial Exercise
A sample space consists of five simple events with P(E₁) = P(E₂) = 0.3, P(E3) = 0.1, and P(E4) = 2P(E5).
Find the probabilities for simple events E4 and E5.
Step 1
Recall that the sum of the probabilities for all simple events in a sample space equals 1. Since there are five simple events, E₁, E₂, E3, E4, E5, this yields the following equation.
P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1
We are given that P(E₁) = P(E₂) = 0.3 and P(E3) = 0.1. Substitute these values into the above equation and simplify.
P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1
0.3 +0.3 +
+ P(E4) + P(E5)
+ P(E4) + P(E5) = 1
P(E4) + P(E5)
=
=
1
Expert Solution
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Follow-up Question

Transcribed Image Text:Step 2
We have simplified P(E₁) + P(E₂) + P(E3) + P(E4) + P(E5) = 1 to the equation P(E4) + P(E5) = 0.3. We are also given that P(E4) = 2P(E5). Note that both of these equations involve only P(E4) and P(E5). Substitute the second equation into
the first equation to solve for P(E5).
P(E4) + P(E5) = 0.3
2P(E5) + P(E5) = 0.3
P(E5) = 0.3
P(E5)
=
Solution
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