Probability of malaria

A survey carried out at a certain hospital indicates that the probability that a patient tested positive for malaria is 0.6. What is the probability that two patients selected at random
(i) one is negative while the other tested positive?
(i) both patients tested positive.

Result

p1 =  0.48
p2 =  0.36

Solution:

p=0.6  C1(2)=(21)=2!1!(21)!=21=2  p1=(21) p (1p)=2 0.6 (10.6)=1225=0.48p = 0.6 \ \\ \ \\ C_{{ 1}}(2) = \dbinom{ 2}{ 1} = \dfrac{ 2! }{ 1!(2-1)!} = \dfrac{ 2 } { 1 } = 2 \ \\ \ \\ p_{ 1 } = { { 2 } \choose 1 } \cdot \ p \cdot \ (1-p) = 2 \cdot \ 0.6 \cdot \ (1-0.6) = \dfrac{ 12 }{ 25 } = 0.48
p2=p2=0.62=925=0.36p_{ 2 } = p^2 = 0.6^2 = \dfrac{ 9 }{ 25 } = 0.36







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