Answer :
Answer:
0.980
Step-by-step explanation:
The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05
So, probability of success = 0.05
Probability of failure = 1-0.05=0.95
There are 12 amplifiers
We are supposed to find the probability that at most two will exceed 2dB.
We will use binomial distribution
Formula : [tex]P(X=r)=^nC_r p^r q ^{n-r}[/tex]
p = 0.05
q = 0.95
n = 12
We are supposed to find the probability that at most two will exceed 2dB.
So, [tex]P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)[/tex]
[tex]P(X\leq 2)=^{12}C_0 P(0.05)^0 (0.95)^{12-0}+^{12}C_1 P(0.05)^1(0.95)^{12-1}+^{12}C_2 P(0.05)^2 (0.95)^{12-2}[/tex]
[tex]P(X\leq 2)=\frac{12!}{0!(12-0)!} (0.05)^0 (0.95)^{12-0}+\frac{12!}{1!(12-1)!}(0.05)^1(0.95)^{12-1}+\frac{12!}{2!(12-2)!} (0.05)^2 (0.95)^{12-2}[/tex]
[tex]P(X\leq 2)=0.980[/tex]
Hence the probability that at most two will exceed 2dB is 0.980