Question 1175558
E(X)=np=500*0.0046=2.3
Normal approximation would not be too accurate given thet np is not > 10. Also, normality will not apply to this skewed distribution. 
V(X)=np(1-p)=2.3*0.9954=2.289
sd=sqrt(V)=1.51
for no mishandled bags (approximation) it is z=(0-2.289)/1.51 or -1.51 or probability 0.0668. Note: this is z < 0 strictly speaking, but < 0 doesn't exist for lost bags, so the approximation for normality will break down here.
exact value is 0.9954^50=0.0997
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more than 2 approximation is z>(2.5-2.3)/1.51 using the continuity correction factor or z>0.13
that probability is 0.4483
exact value: we know 0, and 1 would be 500*0.0046*0.9954^499=0.2304
2 would be 500C2*0.0046^2*0.9954^498=0.2657
those three probabilities add to 0.5958
The complement, and the answer, is 1-0.5958 or 0.4042