statistics - Conditional Probability in R -

i using prob package in r calculate conditional probability.

my data set

 q1 q2 q3 q4     1  1  0  0     0  0  0  0     0  1  0  1     0  1  0  1   

i want calculate prob(q2 =1 given q4=1), per knowledge should 1. when use following command in r

prob(a,q2==1,q4==1) return 0.5

how come return 0.5? 0.5, right? doubting answer.

the second question if change data set

  q1 q2 q3 q4    1  1  0  0    1  0  1  0    0  1  0  1    1  1  1  1  

when use above data , calculate above probability returns 1. how come probability changes when not changing q2 , q4.
thinking should same 1 in both cases.

how come changes change in other parameter q1 , q3. think should change p(q2=1 / q4=1) independent of q1 , q3.

the problem prob uses intersect excludes duplicates. calculation sum(intersect(a, b)$probs)/sum(b$probs) 0.25/0.5=0.5.

if want correct calculation, have use exclusive probabilities (the 3rd line has probability of 50%):

a <-read.table(text="q1 q2 q3 q4   1  1  0  0   0  0  0  0   0  1  0  1",header=true,stringsasfactors=false) a$probs <-c(0.25,0.25,0.5)  prob(a,event=q2==1,given=q4==1) [1] 1 

as second question, prob working correctly because intersect not removing duplicates because line 3 , 4 different.