Longines said:Hey guys,
I don't understand how this question works... I don't understand the answers either. Could someone take me through this step-by-step?
See attached image:
Lol... once again, a simple step that I did not see.chisigma said:a) is...
$\displaystyle P \{ G = k\} = \int_{k-1}^{k} e^{- \lambda\ x}\ d x = e^{\lambda\ k} (e^{\lambda} - 1)\ (1)$
b) for the Bayes theorem is...
$\displaystyle P\{X > k + x| G > k \} = \frac{P \{ X > k + x \}}{P\{X>k \}} = \frac{e^{- \lambda\ (k + x)}}{e^{- \lambda\ k}} = e^{- \lambda\ x}\ (2) $
Kind regards
$\chi$ $\sigma$