Question

It is known that 20% of all persons given a certain sedative drug get drowsy within 2 minutes. Find the probabilities that among 14 persons given this drug.

a)   At most 2          b) at least 5    c) 2,3,4 will get very drowsy within 2 minutes.

Answer 1

X~Bin(14, 0.2)

a) $P(X⩽2)=P(X=0)+P(X=1)+P(X=2)={14 \choose 0}*0.2^0*0.8^{14}+{14 \choose 1}*0.2^1*0.8^{13}+{14 \choose 2}*0.2^2*0.8^{12}=0.044+0.154+0.25=0.448$

b) $P(X⩾5)=1-P(X<5)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)=1-0.044-0.154-0.25-0.25-0.172=0.13$

c) All of the needed probabilities were found in the part (a) and (b)

$P(X=2)=0.25$

$P(X=3)=0.25$

$P(X=4)=0.172$