A manufacturer of digital phones has the following probability distribution for the number of defects per phone: xf(x) .89 1.07 2.03 3.01 (a) Determine the probability of 2 or more defects. (b) Is a randomly selected phone more likely to have 0 defects or 1 or more defects?

A manufacturer of digital phones has the following probability distribution for the number of defects per phone: xf(x) .89 1.07 2.03 3.01 (a) Determine the probability of 2 or more defects. (b) Is a randomly selected phone more likely to have 0 defects or 1 or more defects?

Question
Probability
asked 2020-10-19
A manufacturer of digital phones has the following probability distribution for the number of defects per phone: xf(x) .89 1.07 2.03 3.01
(a) Determine the probability of 2 or more defects. (b) Is a randomly selected phone more likely to have 0 defects or 1 or more defects?

Answers (1)

2020-10-20
a. The probability of 2 or more defects if
P(2 or more defects)=P(2 defects)+P(3 defects),
since one item cannot habe both 2 and 3 defects. So,
P(2 or more defects)=f(2)+f(3)=0.4
b.Similarly,
P(1 or more defects)=f(1)+f(2)+f(3)=.11
Therefore, a phone is more likely to have 0 defects.
0

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