Answer to Question #108806 in Statistics and Probability for Jasmine

Question

A sample of 200 persons is asked about their handedness. A two way table of observed counts follow

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Left-handed & Right-handed& Total\\ \hline Men & 11 &79 & 90 \\ \hdashline Women & 9 & 101 & 110 \\ \hdashline Total & 20 & 180 & 200 \end{array}$$

Let $M=$ selected person is a men, $W=$ selected person is a women, $L=$ selected person is left-handed, $R=$ selected person is right-handed.

If one person is randomly selected

a. $P(W)=\dfrac{110}{200}=\dfrac{11}{20}=0.55$

b. $P(R)=\dfrac{180}{200}=\dfrac{9}{10}=0.9$

c. $P(M\cap R)=\dfrac{79}{200}=0.395$

d. $P(W\cup L)=\dfrac{20+110-9}{200}=\dfrac{121}{200}=0.605$

e. $P(M|L)=\dfrac{11}{20}=0.55$

f. $P(R|W)=\dfrac{101}{110}\approx0.9182$

If one person is randomly selected with replacement

$$P(first\ M\cap L, \ second\ M\cap R )=\dfrac{11}{200}\cdot\dfrac{79}{200}=$$ $$=\dfrac{869}{40000}=0.021725$$

$$P(first\ W\cap L, \ second\ W\cap L )=\dfrac{9}{200}\cdot\dfrac{9}{200}=$$ $$=\dfrac{81}{40000}=0.002025$$

If one person is randomly selected without replacement

$$P(first\ M\cap L, \ second\ M\cap R )=\dfrac{11}{200}\cdot\dfrac{79}{199}=$$ $$=\dfrac{869}{39800}\approx0.021834$$

$$P(first\ W\cap L, \ second\ W\cap L )=\dfrac{9}{200}\cdot\dfrac{9}{199}=$$ $$=\dfrac{81}{39800}\approx0.002035$$