Q:

A student must choose exactly two out of three electives: art, French, and mathematics. He chooses art with probability 5/8, French with probability 5/8, and art and French together with probability 1/4. What is the probability that he chooses mathematics? What is the probability that he chooses either art or French?

Accepted Solution

A:
Answer:The probability of choosing mathematics =P(M)=[tex]\frac{3}{4}[/tex] The probability that he chooses either art or French=1Step-by-step explanation:We are given that a student must choose exactly two out of three elective  subjects : art ,french and mathematics.The probability of choosing art=[tex]\frac{5}{8}[/tex]The  probability of choosing french =[tex]\frac{5}{8}[/tex]The probability of choosing French and art=[tex]\frac{1}{4}[/tex]Let A ,F and M denotes the students of art,french and mathematics.[tex] P(A)=P(A\cap M)+P(A\cap F)[/tex][tex] P(A\cap F)+P(A\cap M)=\frac{5}{8}[/tex][tex] P(F\cap M)+P(F\cap A)=P(F)=\frac{5}{8}[/tex]Probability of choosing mathematics only=0Probability of choosing French only =0Probability of choosing art only =0Probability of choosing all three subjects =0[tex]P(M)=P(M\cap A)+P(M\cap F)[/tex][tex]P(A\cap F)=\frac{1}{4}[/tex]Substitute the value then we get [tex] P(A\cap M)=\frac{5}{8}-\frac{1}{4}==\frac{3}{8}[/tex][tex] P(F\cap M)=\frac{5}{8}-\frac{1}{4}=\frac{3}{8}[/tex]Therefore,[tex] P(M)=P(A\cap M)+P(F\cap M)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}[/tex]Hence, the probability of choosing mathematics =P(M)=[tex]\frac{3}{4}[/tex][tex] P(A\cup F)=P(A)+P(F)-P(A\cap F)[/tex][tex]P(A\cup F)=\frac{5}{8}+\frac{5}{8}-\frac{1}{4}[/tex][tex] P(A\cup F)=\frac{5+5-2}{8}=1[/tex]Hence, the probability that he chooses either art or French=1