PROBLEM 25: A six-sided die is weighted so that Pr[1] Pr[2] Pr[3] = Pr[5], Pr[4] = Pr[6], and Pr[6] =2-Pr[1]. If the die is rolled one time, what is the probability that an even number is rolled? A) 1/8 B) 2/5 C) 1/10 D) 5/8 E) none of the above
PROBLEM 25: A six-sided die is weighted so that Pr[1] Pr[2] Pr[3] = Pr[5], Pr[4] = Pr[6], and Pr[6] =2-Pr[1]. If the die is rolled one time, what is the probability that an even number is rolled? A) 1/8 B) 2/5 C) 1/10 D) 5/8 E) none of the above
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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![**Problem 25**: A six-sided die is weighted so that Pr[1] = Pr[2] = Pr[3] = Pr[5], Pr[4] = Pr[6], and Pr[6] = 2·Pr[1]. If the die is rolled one time, what is the probability that an even number is rolled?
A) 1/8
B) 2/5
C) 1/10
D) 5/8
E) none of the above
To find the probability of rolling an even number, we first acknowledge the probabilities:
1. Let Pr[1] = x. Then Pr[2] = x, Pr[3] = x, and Pr[5] = x.
2. Pr[4] = Pr[6], and Pr[6] = 2x, so Pr[4] = 2x.
3. The sum of all probabilities should equal 1:
\[
x + x + x + x + 2x + 2x = 8x = 1
\]
4. Solving for x gives \( x = \frac{1}{8} \).
Using \( x = \frac{1}{8} \):
- Pr[1] = Pr[2] = Pr[3] = Pr[5] = \(\frac{1}{8}\)
- Pr[4] = Pr[6] = 2x = \(\frac{2}{8} = \frac{1}{4}\)
The probability of rolling an even number (2, 4, 6) is:
\[
Pr[2] + Pr[4] + Pr[6] = \frac{1}{8} + \frac{1}{4} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} + \frac{2}{8} = \frac{5}{8}
\]
Therefore, the answer is D) 5/8.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85ac4c2f-7d2a-4117-b171-587c30248ab0%2F834cf58e-0402-44f3-b71b-255968cbf0c2%2Fbz02b5d_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 25**: A six-sided die is weighted so that Pr[1] = Pr[2] = Pr[3] = Pr[5], Pr[4] = Pr[6], and Pr[6] = 2·Pr[1]. If the die is rolled one time, what is the probability that an even number is rolled?
A) 1/8
B) 2/5
C) 1/10
D) 5/8
E) none of the above
To find the probability of rolling an even number, we first acknowledge the probabilities:
1. Let Pr[1] = x. Then Pr[2] = x, Pr[3] = x, and Pr[5] = x.
2. Pr[4] = Pr[6], and Pr[6] = 2x, so Pr[4] = 2x.
3. The sum of all probabilities should equal 1:
\[
x + x + x + x + 2x + 2x = 8x = 1
\]
4. Solving for x gives \( x = \frac{1}{8} \).
Using \( x = \frac{1}{8} \):
- Pr[1] = Pr[2] = Pr[3] = Pr[5] = \(\frac{1}{8}\)
- Pr[4] = Pr[6] = 2x = \(\frac{2}{8} = \frac{1}{4}\)
The probability of rolling an even number (2, 4, 6) is:
\[
Pr[2] + Pr[4] + Pr[6] = \frac{1}{8} + \frac{1}{4} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} + \frac{2}{8} = \frac{5}{8}
\]
Therefore, the answer is D) 5/8.
Expert Solution
Step 1
Pr(1)= Pr(2)=Pr(3)=Pr(5) = X
Pr(4)=Pr(6)= (2*Pr(1)) = 2X
Probability= favorable/total
From addition rule ,
Sum of all probabilities= 1
Pr(1)+Pr(2)+Pr(3)+Pr(4)+Pr(5)+Pr(6) = 1
X + X + X + 2X +X + 2 X = 8X = 1
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