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A challenge .....statistics helP??

A garage repair service is called out to incidents that it classifies as either breakdowns or accidents. The number of breakdowns that is called out to in a day follows a Poisson distribution with mean 4. The number of accidents, independently of the number of the breakdowns that is called out to in a day follows a Poisson distribution with mean 0.8.


Assuming that the number of times the repair service is called out on a day follows a Poisson distribution with mean 4.8. The prob. that the service is called out on exactly 6 occasions in a day is 0.140.


Given that on a particular day the service is called out on exactly 6 occasions, HOW to find the prob that more call-outs are for Breakdowns than for Accidents.
I HAVE LISTED ALL THE POSSIBILITIES: 4B AND 2A , 5B AND 1A , 6B AND 0A. I obtained 0.131 as ans. but it says 0.938.


PLEASE HELP. THANKS.
It's been a while since I've done AS statistics and I no longer have the formula booklet but did you remember that P(B given that A has happened)=P(AnB)divided by P(A) and that the question says "Given that on a particular day the service is called out on exactly 6 occasions".
P(B∣A)=P(BnA)P(A)P(B|A)=\frac{P(BnA)}{P(A)}

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