p = 0.20
q= 1-0.20 =0.80
n = 100
X~B(n=100,p=0.20)
P(X≤10)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)P(X=x)=(x100)pxq100−xP(X=0)=(0100)p0q100−0=1×1×0.8100=2.037×10−10P(X=1)=(1100)p1q100−1=1!(100−1)!100!×0.21×0.8100−1=100×0.2×0.899=5.092×10−9P(X=2)=(2100)p2q100−2=2!(100−2)!100!×0.22×0.8100−2=299×100×0.22×0.898=6.302×10−8P(X=3)=(3100)p3q100−3=3!(100−3)!100!×0.23×0.8100−3=2×398×99×100×0.23×0.897=5.146×10−7P(X=4)=(4100)p4q100−4=4!(100−4)!100!×0.24×0.8100−4=2×3×497×98×99×100×0.24×0.896=3.120×10−6P(X=5)=(5100)p5q100−5=5!(100−5)!100!×0.25×0.8100−5=2×3×4×596×97×98×99×100×0.25×0.895=1.497×10−5P(X=6)=(6100)p6q100−6=6!(100−6)!100!×0.26×0.8100−6=2×3×4×5×695×96×97×98×99×100×0.26×0.894=2.928×10−5P(X=7)=(7100)p7q100−7=7!(100−7)!100!×0.27×0.8100−7=2×3×4×5×6×794×95×96×97×98×99×100×0.27×0.893=0.000199P(X=8)=(8100)p8q100−8=8!(100−8)!100!×0.28×0.8100−8=2×3×4×5×6×7×893×94×95×96×97×98×99×100×0.28×0.892=0.000578P(X=9)=(9100)p9q100−9=9!(100−9)!100!×0.29×0.8100−9=2×3×4×5×6×7×8×992×93×94×95×96×97×98×99×100×0.29×0.891=0.001478P(X=10)=(10100)p10q100−10=10!(100−10)!100!×0.210×0.8100−10=2×3×4×5×6×7×8×9×1091×92×93×94×95×96×97×98×99×100×0.210×0.890=0.003362P(X≤10)=0.005696
0.57 % is the chance that not more than ten people suffer from COVID 19-disease.
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