Question

At the University of Washington, 65% of the students have been granted a scholarship. Of the scholarship students, 80% also have a work-study job. Of students who did not receive a scholarship, 45% have a work-study job. What is the probability that a student has a scholarship and a work-study job? If a student does not have a work-study job, what is the probability that he received a scholarship?

Answer 1

X=student who has been granted a scholarship

Y=student  has a work-study job

according to the description,

$P(X)=0.65\\ P(X')=1-P(X)=0.35 \\ P(Y|X)=0.80\\ P(Y'|X)=1-P(Y|X)=0.20\\ P(Y|X')=0.45 \\ P(Y'|X')=1-P(Y|X')=0.55\\$

probability that a student has a scholarship and a work-study job=$P(X \cap Y)$

$P(X \cap Y)=P(Y|X)*P(X)=0.80*0.65=0.52$

probability that a student has a scholarship and a work-study job=0.52

 If a student does not have a work-study job,probability

that he received a scholarship= $P(X|Y')$

$P(X|Y')=\frac{P(X\cap Y')}{P(Y')}\quad (eq:A)$

so,

$P(X\cap Y')=P(Y'|X)*P(X)\\ P(X\cap Y')=0.2*0.65=0.13$

and using conditional probability equation,

$P(X|Y')*P(Y')=P(Y'|X)*P(X)\quad(eq:1)\\ P(X'|Y')*P(Y')=P(Y'|X')*P(X')\quad(eq:2)\\$

$(eq:1 +eq:2)\\ P(X|Y')*P(Y')+P(X'|Y')*P(Y')=\\ \hspace{5 em}P(Y'|X)*P(X)+P(Y'|X')*P(X')\\ (P(X|Y')+P(X'|Y'))*P(Y')=\\ \hspace{5 em}P(Y'|X)*P(X)+P(Y'|X')*P(X')\\$

since,$P(X|Y')+P(X'|Y')=1$

$P(Y')=P(Y'|X)*P(X)+P(Y'|X')*P(X')\\ P(Y')=0.2*0.65+0.55*0.35=0.3225\\$

therefore from (eq: A),

$P(X|Y')=\frac{P(X\cap Y')}{P(Y')}\\ P(X|Y')=\frac{0.13}{0.3225}=0.403$

 If a student does not have a work-study job,probability

that he received a scholarship=0.403