Answer:
use photomath!
Step-by-step explanation:
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I believe the second one : drawing a number from a hat, not replacing it, and then drawing a second number
X is greater than or equal to negative 1 and you would right 1 comma infinity
Answer:
For first lamp ; The resultant probability is 0.703
For both lamps; The resultant probability is 0.3614
Step-by-step explanation:
Let X be the lifetime hours of two bulbs
X∼exp(1/1400)
f(x)=1/1400e−1/1400x
P(X<x)=1−e−1/1400x
X∼exp(1/1400)
f(x)=1/1400 e−1/1400x
P(X<x)=1−e−1/1400x
The probability that both of the lamp bulbs fail within 1700 hours is calculated below,
P(X≤1700)=1−e−1/1400×1700
=1−e−1.21=0.703
The resultant probability is 0.703
Let Y be a lifetime of another lamp two bulbs
Then the Z = X + Y will follow gamma distribution that is,
X+Y=Z∼gamma(2,1/1400)
2λZ∼
X+Y=Z∼gamma(2,1/1400)
2λZ∼χ2α2
The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,
P(Z≤1700)=P(1/700Z≤1.67)=
P(χ24≤1.67)=0.3614
The resultant probability is 0.3614
