From noon until sunset, bees emerge from a beehive according to a Poisson proces

Question

From noon until sunset, bees emerge from a beehive according to a Poisson process with rate =4=4 per second.
Find the probability that bees 1 and 2 emerge less than 0.2 seconds apart while bees 3 and 4 emerge more than 0.2 seconds apart.  
Find the probability that bees 5 and 6 emerge less than 0.2 seconds apart while bees 7 and 10 emerge more than 0.2 seconds apart.  
Find the probability that the 2nd bee emerges less than 0.53 seconds after noon and the 4rd bee emerges more than 0.53 seconds after noon. (Hint: Are the events independent?)

I know the answer to the first part is .24743, but the method that I have been trying to use is not getting me this answer.

Explanation / Answer

1)probability that bees 1 and 2 emerge less than 0.2 seconds apart while bees 3 and 4 emerge more than 0.2 seconds apart. =(1-e-t)*e-t =(1-e-0.2*4)*e-0.2*4 =0.247432

2) here first one is exponential distribution with P(X<=0.2)=(1-e-0.2*4)

and second one is poisson distribution with mean (0.2*4) P(X<=2)

hence probability that bees 5 and 6 emerge less than 0.2 seconds apart while bees 7 and 10 emerge more than 0.2 seconds apart =0.550671*0.952577 =0.524557

3)probability that the 2nd bee emerges less than 0.53 seconds after noon and the 4rd bee emerges more than 0.53 seconds after noon. =P(2<=X<=3) =0.460348 (it is a poisson process with mean =0.53*4)

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