I remember this one from a youth mathematics book. 10⁻⁴ chance of one bomb, and 10⁻⁸ of two (assuming the other one is an independent event — which it would be, as you're certainly not in collusion with any actual bomber).
If you always have a bomb with you, then the probability of you sitting on a plane with a bomb is 1, and thus the probability of you sitting on a plane with another bomb is again 10^-4.