Question

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?

Answer 1

Answer:

The probability of choosing mathematics =P(M)=$\frac{3}{4}$

The probability that he chooses either art or French=1

Step-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=$\frac{5}{8}$

The  probability of choosing french =$\frac{5}{8}$

The probability of choosing French and art=$\frac{1}{4}$

Let A ,F and M denotes the students of art,french and mathematics.

$P(A)=P(A\cap M)+P(A\cap F)$

$P(A\cap F)+P(A\cap M)=\frac{5}{8}$

$P(F\cap M)+P(F\cap A)=P(F)=\frac{5}{8}$

Probability of choosing mathematics only=0

Probability of choosing French only =0

Probability of choosing art only =0

Probability of choosing all three subjects =0

$P(M)=P(M\cap A)+P(M\cap F)$

$P(A\cap F)=\frac{1}{4}$

Substitute the value then we get

$P(A\cap M)=\frac{5}{8}-\frac{1}{4}==\frac{3}{8}$

$P(F\cap M)=\frac{5}{8}-\frac{1}{4}=\frac{3}{8}$

Therefore,$P(M)=P(A\cap M)+P(F\cap M)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}$

Hence, the probability of choosing mathematics =P(M)=$\frac{3}{4}$

$P(A\cup F)=P(A)+P(F)-P(A\cap F)$

$P(A\cup F)=\frac{5}{8}+\frac{5}{8}-\frac{1}{4}$

$P(A\cup F)=\frac{5+5-2}{8}=1$

Hence, the probability that he chooses either art or French=1