Answer:
Step-by-step explanation:
We need to find the conditional probability P( T1 < s|N(t)=1 ) for all s ≥ 0
P( time of the first person's arrival < s till time t exactly 1 person has arrived )
= P( time of the first person's arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )
{ As till time t, we know that exactly 1 person has arrived, thus relevant values of s : 0 < s < t }
P( time of the first person arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )
= P( exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )
P(exactly x person has arrived till time t ) ~ Poisson(kt) where k = lambda
Therefore,
P(exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )
= [ kse-ks/1! ] / [ kte-kt/1! ]
= (s/t)e-k(s-t)
Answer:
11
Step-by-step explanation:
Answer:
cross multiply and they should equal the same
Step-by-step explanation:
Answer: I think it should be a: 1/4 inch
Step-by-step explanation:
So first convert to improper fractions
7 and 7/8=7+7/8=56/8+7/8=63/8
3 and 1/4=3+1/4=12/4+1/4=13/4
so nowe we have
63/8+13/4
convert bottom number to same
which is 8
13/4 times 2/2=26/8
now we have
63/8-26/8=37/8=4 and 5/8
answer is 4 and 5/8
