Hello,
Answer A
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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)
The answer is -23,-19,-8,5,14
4 goes into 4 once, and then into 16 4 times. Making the final answer 1/4
