Answer to Question #114336 in Statistics and Probability for kelly

Question #114336
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?
1
Expert's answer
2020-05-08T19:32:35-0400

X=student who has been granted a scholarship

Y=student  has a work-study job


according to the description,

"P(X)=0.65\\\\\nP(X')=1-P(X)=0.35 \\\\\nP(Y|X)=0.80\\\\\nP(Y'|X)=1-P(Y|X)=0.20\\\\\nP(Y|X')=0.45 \\\\\nP(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)\\\\\nP(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)\\\\\nP(X'|Y')*P(Y')=P(Y'|X')*P(X')\\quad(eq:2)\\\\"


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



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