Answer to Question #87276 in Statistics and Probability for Wawan

Question

Mathematics male:8 , female:5
Science male:6, female : 7
English male: 4 , female : 8
History male : 3 , female : 5
If a teacher is selected randomly, calcute these probabilities,
I)mathematics teacher or female
II)history teacher or male
III)not an english teacher
IV)male science teacher
V)mathematics or science teacher

Answer 1

Let M be event that teacher is mathematics teacher, H be the event that teacher is history teacher, E be the event that the teacher is English teacher, and S be the event that teacher is the science teacher.

$$8+5+6+7+4+8+3+5=46$$

$$P(M)={{8+5} \over 46}={13 \over 46}$$

$$P(S)={{6+7} \over 46}={13 \over 46}$$

$$P(E)={{4+8} \over 46}={6 \over 23}$$

$$P(H)={{3+5} \over 46}={4 \over 23}$$

$$P(male)={{8+6+4+3} \over 46}={21 \over 46}$$

$$P(female)={{5+7+8+5} \over 46}={25 \over 46}$$

$$I)\ P(M\cup female)=P(M)+P(female)- P(M\cap female)$$

$$P(M\cup female)={13 \over 46}+{25 \over 46}-{5\over 46}={33 \over 46}$$

$$II)\ P(H\cup male)=P(H)+P(male)- P(H\cap male)$$

$$P(H\cup male)={8 \over 46}+{21 \over 46}-{3\over 46}={13 \over 23}$$

$$III)\ P(\bar{E})=1-P(E)$$

$$P(\bar{E})=1-{6 \over 23}={17 \over 23}$$

$$IV)\ P(S\cap male)={6 \over 46}={3 \over 23}$$

$$V)\ P(M\cup S)=P(M)+P(S)$$

$$P(M\cup S)={13 \over 46}+{13 \over 46}={13 \over 23}$$