A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
There are 52 total cards in a deck. The 13th can be any of that 52 cards. There are 4 Ace in a deck so probability of that is \(\frac {4}{52} = \frac{1}{13}\)
The Acme Super light bulb is known to have a useful life described by the density function \[ f(t)=.01e^{-0.1t} \] Where t is measured in hours.
A) Find the failure rate of this bulb (see Exercise 2.2.6)
B) Find the reliability of this bulb after 20 hours.
C) Given that it lasts 20 hours, find the probability that the bulb lasts another 20 hours.
D) Find the probability that the bulb burns out in the forty-first hour, given that it lasts 40 hours.
A)The failure rate of the bulb is \[ -\frac{R'(t)}{R(t)} \]
\[ -\frac{R'(t)}{R(t)}=-\frac{-0.01e^{-0.01t}}{e^{0.01t}}=0.01 \]
\[ \begin{aligned} P&=\frac{R(40)}{R(20)}\\ &=\frac{e^{-0.01(40)}}{e^{-0.01(20)}}\\ &=0.82 \end{aligned} \] D) In order to compute this probability we need to consider the numerator as the difference between the failure functions F(41)-F(40). The denominator will be the realiability evaluated at 40. Our probability is as follows
\[ \begin{aligned} P&=\frac{F(41)-F(40)}{R(40)}\\ &=\frac{(1-e^{-0.01(41)})-(1- e^{-0.01(40)})}{e^{-0.01(40)}}\\ &=\frac{(e^{-0.01(40)})-e^{-0.01(41)}) }{e^{-0.01(40)}}\\ &=0.01 \end{aligned} \]