QUESTION IMAGE
Question
finding the probabilities of multiple events
defective parts: x | probability: p(x)
0 | 0.85
1 | 0.10
2 | 0.04
3 | 0.01
shelia does quality control for a company that manufactures lawn mower parts. on any given day, she finds the probability distribution for defective parts as shown in the table.
using the data from the table, what is the probability of having 2 defective parts in a day?
what is the probability that on a given day there, shelia will find defects?
First Sub - Question: Probability of 2 defective parts
Step1: Identify the relevant probability
We look at the table where the number of defective parts \(X = 2\), and the corresponding probability \(P(X)\) is given.
From the table, when \(X = 2\), \(P(X)=0.04\).
Step1: Understand the complement rule
The probability of finding defects is the complement of the probability of finding 0 defective parts. The complement rule states that \(P(A)=1 - P(\text{not }A)\). Here, \(A\) is the event of finding defects, and "not \(A\)" is the event of finding 0 defective parts.
Step2: Calculate the probability
We know that \(P(X = 0)=0.85\). So the probability of finding defects \(P(\text{defects})=1 - P(X = 0)\).
Substitute \(P(X = 0) = 0.85\) into the formula: \(1-0.85 = 0.15\).
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0.04