Question

\[ \begin{aligned} \text { PCACBCCACBCC) } & =\text { AC. } \times C \times C A \times C B \times C C \\ & =0.0234 \times 0.0138 \times 0.0234 \times 0 . \\ & =1-0.000073263 \\ & =0.99 \end{aligned} \] \[ \begin{aligned} P(A B B A) & =0.0468 \times 0.0468 \\ & =1-0.00219024 \\ & =0.99 \end{aligned} \] \( 0 \leq P(A) \leq \).1 , There is ns zere prowabinility twie ro imposible to get zero probability. \[ \begin{array}{l} \lambda=8 \\ f(10<x<15)=e^{-\lambda a}-e^{-\lambda b} \\ \left.=e^{-8(10}-e^{-86}\right) \\ =e^{-80}-e^{-120} \\ =1.80 \times 10^{-35}-7.69 \times 10^{-53} \\ =1.80 \times 10^{-35} \end{array} \]

          \[
\begin{aligned}
\text { PCACBCCACBCC) } & =\text { AC. } \times C \times C A \times C B \times C C \\
& =0.0234 \times 0.0138 \times 0.0234 \times 0 . \\
& =1-0.000073263 \\
& =0.99
\end{aligned}
\]
\[
\begin{aligned}
P(A B B A) & =0.0468 \times 0.0468 \\
& =1-0.00219024 \\
& =0.99
\end{aligned}
\]
\( 0 \leq P(A) \leq \).1 , There is ns zere prowabinility twie ro imposible to get zero probability.
\[
\begin{array}{l} 
\lambda=8 \\
f(10<x<15)=e^{-\lambda a}-e^{-\lambda b} \\
\left.=e^{-8(10}-e^{-86}\right) \\
=e^{-80}-e^{-120} \\
=1.80 \times 10^{-35}-7.69 \times 10^{-53} \\
=1.80 \times 10^{-35}
\end{array}
\]

Show more…

PCACBCCACBCC) = AC. × C × C A × C B × C C
=0.0234 × 0.0138 × 0.0234 × 0 .
=1-0.000073263
=0.99

P(A B B A) =0.0468 × 0.0468
=1-0.00219024
=0.99

0 ≤ P(A) ≤.1 , There is ns zere prowabinility twie ro imposible to get zero probability.

λ=8

f(10<x<15)=e^-λ a-e^-λ b
.=e^-8(10-e^-86)

=e^-80-e^-120

=1.80 × 10^-35-7.69 × 10^-53

=1.80 × 10^-35

Added by Stacy E.

Advanced Engineering Mathematics

Erwin Kreyszig 9th Edition

Chapter 7

Instant Answer

Solved by Expert Adi S

07/01/2023

Step 1

Step 2: Analyze the first part of the problem: \[ \text{PCACBCCACBCC} = \text{AC} \times C \times CA \times CB \times CC \] \[ = 0.0234 \times 0.0138 \times 0.0234 \times 0 \] \[ = 1 - 0.000073263 \] \[ = 0.99 \] Step 3: Identify the mistake in the calculation. Show more…

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Thanks for your feedback!

\[ \begin{aligned} \text { PCACBCCACBCC) } & =\text { AC. } \times C \times C A \times C B \times C C \\ & =0.0234 \times 0.0138 \times 0.0234 \times 0 . \\ & =1-0.000073263 \\ & =0.99 \end{aligned} \] \[ \begin{aligned} P(A B B A) & =0.0468 \times 0.0468 \\ & =1-0.00219024 \\ & =0.99 \end{aligned} \] \( 0 \leq P(A) \leq \).1 , There is ns zere prowabinility twie ro imposible to get zero probability. \[ \begin{array}{l} \lambda=8 \\ f(10<x<15)=e^{-\lambda a}-e^{-\lambda b} \\ \left.=e^{-8(10}-e^{-86}\right) \\ =e^{-80}-e^{-120} \\ =1.80 \times 10^{-35}-7.69 \times 10^{-53} \\ =1.80 \times 10^{-35} \end{array} \]

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