A survey was conducted to find out the success rate of students who qualified the entrance examination by dropping a year after class XII.
As per the data collected, 40% students appearing in the examination were dropouts and the remaining students were regular students of class XII. Of the dropouts, 5% qualify the examination while 10% of the regular students qualify the examination.
Find the probability that a student selected at random is a regular
student.
Solution and Explanation
Given:
- Probability of selecting a dropout student, \[ P(D)=40\%=0.4 \]
- Therefore, probability of selecting a regular student, \[ P(R)=60\%=0.6 \]
Hence,
\[ \boxed{P(R)=0.6=\frac{3}{5}} \]
Answer: \(\boxed{\frac{3}{5}}\)
What is the probability that the student will not qualify the
examination ?
Solution and Explanation
Given:
- Among dropout students, \[ P(Q|D)=5\%=0.05 \]
- Therefore, \[ P(Q'|D)=1-0.05=0.95 \]
Hence, the required probability is
\[ \boxed{P(Q'|D)=0.95=\frac{19}{20}} \]
Answer: \(\boxed{\frac{19}{20}}\)
A student selected at random qualified the examination. Find
the probability that student is not a dropout.
Solution and Explanation
Given:
- \[ P(D)=0.4,\qquad P(R)=0.6 \]
- \[ P(Q|D)=0.05,\qquad P(Q|R)=0.10 \]
Step 1: Find the probability of qualifying.
\[ \begin{aligned} P(Q) & amp;=P(D)P(Q|D)+P(R)P(Q|R)\\ & amp;=0.4(0.05)+0.6(0.10)\\ & amp;=0.02+0.06\\ & amp;=0.08 \end{aligned} \]
Step 2: Apply Bayes' theorem.
\[ \begin{aligned} P(R|Q) & amp;=\frac{P(R)P(Q|R)}{P(Q)}\\ & amp;=\frac{0.6\times0.10}{0.08}\\ & amp;=\frac{0.06}{0.08}\\ & amp;=\frac{3}{4} \end{aligned} \]
Answer:
\[ \boxed{\frac{3}{4}} \]
A student selected at random did not qualify the examination.
Find the probability that the student was a regular student.
Solution and Explanation
Given:
- \[ P(D)=0.4,\qquad P(R)=0.6 \]
- \[ P(Q|D)=0.05,\qquad P(Q|R)=0.10 \]
Step 1: Find the probability of not qualifying.
\[ \begin{aligned} P(Q') & amp;=P(D)P(Q'|D)+P(R)P(Q'|R)\\ & amp;=0.4(0.95)+0.6(0.90)\\ & amp;=0.38+0.54\\ & amp;=0.92 \end{aligned} \]
Step 2: Apply Bayes' theorem.
\[ \begin{aligned} P(R|Q') & amp;=\frac{P(R)P(Q'|R)}{P(Q')}\\ & amp;=\frac{0.6\times0.90}{0.92}\\ & amp;=\frac{0.54}{0.92}\\ & amp;=\frac{27}{46} \end{aligned} \]
Answer:
\[ \boxed{\frac{27}{46}} \]
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