The figure shows a one-way road network system from Town P to Towns R, S and T. Any car leaving Town P will pass through either Tunnel A or Tunnel B and arrive at Towns R, S and T via the roundabout Q. A survey shows that 2/5 of the cars leaving P will pass through Tunnel A. The survey also shows that 1/7 of all the cars passing through the roundabout Q will arrive at R, 2/7 at S and 4/7 at T. (a) Two cars leave P. Find the probabilities that: (i) one of them arrives at R and the other one at S, (ii) both of them arrive at S, one through Tunnel A and the other one through Tunnel B.
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@phi @ganeshie8
a)one of them arrives at R and the other one at S, p(R)*p(S)+p(S)*P(R) 1/7 *2/7+2/7*1/7=4/49
does it also include the tunnels?
hmmm let me check again
sry i guess im wrong , u shud check all the possiple tunnels too
the book's answer is 4/49. But i don't know is it necessary to calculate the tunnels or not..
why we don't need to calculate the tunnels?
i guess its not necessary , but u need to check from somone else
ok, and part b?
the tunnels do not matter... it is 100% chance that you go from P to Q if you don't care how you get to Q, the tunnels do not matter.
@phi oh i see...
hmm this is wat i cud think of:o ok so they arrive P p(A)p(s)*P(B)P(S)+P(B)P(S)*p(A)p(s) (2/5*2/7)*(1/5*2/7)+(1/5*2/7)+(2/5*2/7)=16/1225
for (b), there are 4 ways: AA, AB, BA, BB to get to Q and then 4/49 chance from Q to SS
@ikram002p i also do it like your way, but the answer is not 16/1225... @phi yes....
aha ic hmm wats te correct answer :o
ikram should use 2/5 and 3/5 (not 1/5
ops!
@ikram002p @phi oh i got it now, i also made the same mistake from @ikram002p .... the answer is 48/1225?
yes, that is what I got 48/1225
hmm so this is how we got the answer ? p(A)p(s)*P(B)P(S)+P(B)P(S)*p(A)p(s) (2/5*2/7)*(3/5*2/7)+(3/5*2/7)+(2/5*2/7)=48/1225
@ikram002p yes.
@phi @ikram002p thank you both :D
np :) im not good in this lol
@ikram002p haha i still have another questions ...
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