The appropriate distribution for this question is Poisson distribution.
"P(X=k) = \\frac{\u03bb^k \\times e^{-\u03bb}}{k!} \\\\\n\nP(X\u226520) = 1 -P(X<20) \\\\\n\n= 1 - \\sum^{19}_{k=0}P(\u03bb,k)"
Mean λ=13
By Excel function
P(X<20) = POISSON(19,13,1)
= 0.9573
P(X≥20) = 1 -0.9573 = 0.0427
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