Question

Supposewehavealatenightbusandtowardstheendofthe route, there are 3 passengers {P1 , P2 ,, P3} and 5 stops {S1,S2,S3,S4,S5, } remain. Suppose further that each passenger is inebriated, and is thus is equally likely to get off at any one of the stops.

(i)

We wish to list the set of outcomes in the sample space each of whose outcomes is an ordered triple of all three Sij for I=1,2,3, where Sij means that passenger Pi got off at the stop Sj.

1. a) Write the outcome that P1 and P2 debarked at S4 , and P3 debarked at S1. HINT: Its form is (--,--,--) where each – is an integer from 1 to 5.

2. b) How many outcomes are in our sample space?

3. (ii) Using (i), find the probability that no two of our passengers got off at the same stop.

4. There will definitely be a fight if exactly two passengers get off at any given stop. (If nobody or 1 passenger gets off at a stop, there is nobody to fight with, and if all three get off at the same stop, one guy will talk the other two out of fighting.) So what is the probability of a fight occurring?

5. Finally, suppose there are n stops remaining, and k< n passengers. Answer question (ii) now.

Answer 1

$(i)~(a)~ (S_{14},S_{24},S_{31})\\ (b)~Total~no.~of~outcomes~in~sample~space=5\times 5\times 5=5^3=125\\ (ii)~Since~no~two~of~our~passengers~got~off~at~the~same~stop~then~for~first~passenger~\\ we~have~5~stops~i.e.~Passenger~1~gets~off~in~any~one~of~5~stops~and~then~for~Passenger~2\\ we~have~4~stops~and~for~Passenger~3~we~have~remaining~3~stops.\\~Hence~total~no.~of~cases~favourable~to~the~event~that\\ no~two~of~our~passengers~got~off~at~the~same~stop=5\times 4\times 3\\ Hence~required~probability~that~no~two~of~our~passengers~got~off~at~the~same~stop\\ =\frac{5\times 4\times 3}{5^3}=\frac{12}{5^2}=0.48\\ ~Probability~of~a~fight~occurring=1-probability~that~no~two~of~our~passengers~got~off~at~the~same~stop\\ =1-0.48=0.52\\ ~Probability~that~no~two~of~our~passengers~got~off~at~the~same~stop\\ =\frac{n(n-1)...(n-k+1)}{n^k}=\frac{(n-1)...(n-k+1)}{n^{k-1}}\\$

$Probability~of~a~fight~occurring=1-probability~that~no~two~of~our~passengers~got~off~at~the~same~stop\\ =1-\frac{n(n-1)...(n-k+1)}{n^k}=1-\frac{(n-1)...(n-k+1)}{n^{k-1}}$

Now for Question (i), n5, k 3 then Probability that no two of 12 0.48 25 Probability of a fighting occurring stop our passengers got off at the same 4 x 3 52 1 -0.48 0.52 OC