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Chapter 23: Problem 3

Components are made by machines A, B and C. Machine A makes \(35 \%\) of the components, machine B makes \(25 \%\) and machine C makes the rest. Two components are picked at random. Calculate the probability that (a) both are made by machine \(\mathrm{C}\) (b) one is made by machine A and one is made by machine B (c) exactly one is made by machine \(\mathrm{A}\) (d) at least one is made by machine B (e) both are made by the same machine.

Short Answer

Expert verified
Question: Calculate the following probabilities when picking two components made by machines A, B, and C: a) Both components are made by machine C. b) One component is made by machine A, and the other is made by machine B. c) Exactly one component is made by machine A. d) At least one component is made by machine B. e) Both components are made by the same machine. Answer: a) 0.159 b) 0.171 c) 0.277 d) 0.719 e) 0.342

Step by step solution

01

Determine Probabilities for A, B, and C

Based on the percentages given in the question, we know that Machine A makes 35% of the components, B makes 25%, and C makes the rest. Assign probabilities to each machine: P(A) = \(0.35\) P(B) = \(0.25\) P(C) = \(1 - (P(A)+P(B)) = 1 - (0.35+0.25) = 0.40\)
02

- Calculate Probability of both being made by machine C

To calculate the probability of both components being made by machine C, we will use the following formula: P(C and C) = P(C) * P(C|C) P(C|C) represents the probability of the second component being made by machine C, given that the first component was also made by machine C. Since there are fewer components left, P(C|C) = \((0.40 * \frac{0.40-1}{0.40}) = 0.396\) Now we can find the probability of both being made by Machine C: P(C and C) = 0.40 * 0.396 ≈ 0.159
03

- Calculate Probability of one being made by machine A and one by machine B

The components can be selected in two ways: AB or BA. We need to calculate the probabilities for both scenarios and then add them: P(AB) = P(A) * P(B|A) P(BA) = P(B) * P(A|B) P(B|A) = \(0.25 * \frac{0.25-1}{0.25}\) = 0.242 P(A|B) = \(0.35 * \frac{0.35-1}{0.35}\) = 0.342 P(AB) = 0.35 * 0.242 ≈ 0.085 P(BA) = 0.25 * 0.342 ≈ 0.086 Now, to calculate the probability for (b): P(A and B) = P(AB) + P(BA) = 0.085 + 0.086 ≈ 0.171
04

- Calculate Probability of exactly one being made by machine A

The components can be picked in two ways, either AC or CA: P(AC) = P(A) * P(C|A) P(CA) = P(C) * P(A|C) P(C|A) = \(0.40 * \frac{0.40-1}{0.40}\) = 0.391 P(A|C) = \(0.35 * \frac{0.35-1}{0.35}\) = 0.351 P(AC) = 0.35 * 0.391 ≈ 0.137 P(CA) = 0.40 * 0.351 ≈ 0.140 Now, to calculate the probability for (c): P(Exactly one A) = P(AC) + P(CA) = 0.137 + 0.140 ≈ 0.277
05

- Calculate Probability of at least one being made by machine B

First, we will calculate the complementary probability of none being made by machine B, which is a combination of AA and CC: P(None B) = P(AA) + P(CC) P(AA) = P(A) * P(A|A) = 0.35 * 0.349 ≈ 0.122 P(CC) = P(C) * P(C|C) = 0.40 * 0.396 ≈ 0.159 P(None B) = 0.122 + 0.159 ≈ 0.281 Now, to find the probability of at least one B, use the complementary probability: P(At least one B) = 1 - P(None B) = 1 - 0.281 ≈ 0.719
06

- Calculate Probability of both being made by the same machine

The probability that both components are made by the same machine can be calculated by summing the probabilities of AA, BB, and CC: P(Same machine) = P(AA) + P(BB) + P(CC) P(BB) = P(B) * P(B|B) = 0.25 * 0.243 ≈ 0.061 P(Same machine) = 0.122 + 0.061 + 0.159 ≈ 0.342 The probabilities for each part of the exercise are as follows: (a) 0.159 (b) 0.171 (c) 0.277 (d) 0.719 (e) 0.342

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