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.