A man does not have a ticket but still rides the train each weekday (assume a one way ride). Ticket control will get on the train with 0.2 probability each day, and if they do, the man will get caught with 0.95 chance.
i) What is the probability of him not getting caught the entire week (5 days) ?
ii) What is the probability of him getting caught twice?
iii) What is the probability of first getting caught on a Thursday?
$P(A\cap B) = P(A) * P(B) = 0.95 * 0.2 = \frac{19}{100} $ chance of getting caught each day.
i) On five days, its five times the chance, $ 5*\frac{19}{100} = \frac{95}{100}$, so not getting caught is $ 1 - \frac{95}{100} = \frac{5}{100} $
ii) Getting caught on a day needs to happen twice, so $ (\frac{19}{100})^2 = \frac{361}{10000} $
iii) Getting caught after three days of not getting caught means $ (1-\frac{19}{100})^3 + \frac{19}{100} \approx 72\% $
Am I anywhere near right with these answers?
The probability of him getting caught on a given day is $0.2\cdot0.95=0.19$
The probability of him not getting caught on a given day is $1-0.19=0.81$
The probability of him not getting caught during the entire week is $0.81^5\approx35\%$
The probability of him getting caught on a given day is $0.2\cdot0.95=0.19$
The probability of him not getting caught on a given day is $1-0.19=0.81$
The probability of him getting caught exactly twice is $\binom52\cdot0.19^2\cdot0.81^{5-2}\approx19\%$
The probability of him getting caught on a given day is $0.2\cdot0.95=0.19$
The probability of him not getting caught on a given day is $1-0.19=0.81$
The probability of him first getting caught on the fourth day is $0.81^3\cdot0.19\approx10\%$