Question

everyday number of traffic accidents has a distribution x 0,1,2 and more than two having P(x) 0.6,0.2,0.2,0 respectively independent of the other days. what is the probability that there are more accidents on friday than on thursday ?

Answer 1

I don't know... :/ sorrryyy probably more accidents of friday bc of the drunk drivers though lols :p

Answer 2

ok np... thanks anyway :)

Answer 3

I get .28

Answer 4

One way to do this...let X=accidents on friday and Y=accidents on friday

you want P(X>Y)

=P(X=1,Y=0 or X=2,Y=0 or X=2,Y=1)

Answer 5

you can also use symmetry

Answer 6

thanks zarkon

Answer 7

@Zarkon Can you go through the question with Annas rather than just an outline as he is still confused.

Answer 8

Using symmetry...

by symmetry \(P(X>Y)=P(X<Y)\)

using the fact that

\[1=P(X>Y \text{ or }X=Y \text{ or }X<Y)=P(X>Y)+P(X=Y)+P(X<Y)\]
\[=2P(X>Y)+P(X=Y)\]

solving for \(P(X>Y)\)

we get

\[P(X>Y)=\frac{1-P(X=Y)}{2}\]

and \[P(X=Y)=P(X=0,Y=0 \text{ or }X=1,Y=1 \text{ or }X=2,Y=2)\]
\[=P(X=0,Y=0)+P(X=1,Y=1)+P(X=2,Y=2)\]
\[=.6^2+.2^2+.2^2=.44\]

so \[P(X>Y)=\frac{1-P(X=Y)}{2}=\frac{1-.44}{2}=\frac{.56}{2}=.28\]