Question

A JEE aspirant estimates that she will be successful with an 80 percent chance if she studies 10 hours per day, with a 60 percent chance if she studies 7 hours per day and with 40 percent chance if she studies 4 hours per day. She further believes that she will study 10 hours, 7 hours, and 4 hours per day with probabilities 0.1, 0.2 and 0.7 respectively. Given that she does not achieve success, the chance she studied for 4 hour, is\[\]
A.$\dfrac{18}{26}$\[\]
B. $\dfrac{19}{26}$\[\]
C. $\dfrac{20}{26}$\[\]
D. $\dfrac{21}{26}$\[\]

Answer 1

Hint: We denote the event that that the JEE student will study 10 hours , 7 hours and 4 hours per day as ${{E}_{1}},{{E}_{2}},{{E}_{3}}$. We are given the probabilities $P\left( {{E}_{1}} \right),P\left( {{E}_{2}} \right),P\left( {{E}_{3}} \right)$. We denote the new event of not achieving success as $X$. We find the probability that she is not successful subjected to condition that she has studied 10 hours as $P\left( X|{{E}_{1}} \right)$, she has studied 7 hours as $P\left( X|{{E}_{1}} \right)$, she has studied 4 hours as $P\left( X|{{E}_{1}} \right)$. We find the required probability that she has not achieved success subjected to condition that she has studied 4 hours $P\left( {{E}_{3}}|X \right)$ using Bayes’ theorem $P\left( {{E}_{i}}|X \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}}$ for $n=3$. \[\]

Complete step-by-step solution:
Bayes’ theorem is used when there are more than to two events dependent on each other.. If there are $n$ events say ${{E}_{1}},{{E}_{2}},{{E}_{3}},...,{{E}_{n}}\left( {{E}_{i}}\ne \Phi \right)$. Let $X$ be a new event such that $X\subset \bigcup\limits_{i=1}^{n}{{{E}_{i}}},P\left( X \right)\ne 0$, then the probability of event ${{E}_{i}}$ has happened subjected to $X$ has happened is given by

\[P\left( {{E}_{i}}|X \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}}\]
Where $P\left( {{E}_{i}} \right)$ are prior probabilities, $P\left( X|{{E}_{i}} \right)$ are posterior probabilities and $P\left( X \right)$ is the probability of new event. \[\]
 Let us denote the event that that the JEE student will study 10 hours , 7 hours and 4 hours per day as ${{E}_{1}},{{E}_{2}},{{E}_{3}}$. We are given in the question that the probability she will study 10 hours , 7 hours and 4 hours as $P\left( {{E}_{1}} \right)=0.1,P\left( {{E}_{2}} \right)=0.2,P\left( {{E}_{3}} \right)=0.7$ .These are our prior probabilities. \[\]
We denote that she will be successful as $S$ and not successful as $X$. We are also given in the question that that the probability that she will be successful subjected to condition that she studies 10 hours is $P\left( S|{{E}_{1}} \right)=80 \%=0.8$, she studies 7 hours is $P\left( S|{{E}_{2}} \right)=60 \%=0.6$ and she studies 4 hours is $P\left( S|{{E}_{3}} \right)=40 \% =0.4$.\[\]
We are asked to find the probability she has studied four hours after the new event of being not successful which is $P\left( {{E}_{3}}|X \right)$. So we need to find posterior probabilities. The probability that she will not be successful after studying 10 hours is $P\left( X|{{E}_{1}} \right)=1-P\left( S|{{E}_{1}} \right)=1-0.8=0.2$, after studying 7 hours is $P\left( X|{{E}_{2}} \right)=1-P\left( S|{{E}_{2}} \right)=1-0.6=0.4$ and after studying 4 hours is $P\left( X|{{E}_{3}} \right)=1-P\left( S|{{E}_{3}} \right)=1-0.4=0.6$. \[\]
Now we can use the Bayes’ theorem for three events to find $P\left( {{E}_{3}}|X \right)$
\[\begin{align}
  & P\left( {{E}_{3}}|X \right)=\dfrac{P\left( {{E}_{3}} \right)P\left( X|{{E}_{3}} \right)}{\sum\limits_{i=1}^{3}{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}} \\
 & \Rightarrow P\left( {{E}_{3}}|X \right)=\dfrac{P\left( {{E}_{3}} \right)P\left( X|{{E}_{3}} \right)}{P\left( {{E}_{1}} \right)P\left( X|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( X|{{E}_{2}} \right)+P\left( {{E}_{3}} \right)P\left( X|{{E}_{3}} \right)} \\
 & \Rightarrow P\left( {{E}_{3}}|X \right)=\dfrac{0.7\times 0.6}{0.1\times 0.2+0.2\times 0.4+0.7\times 0.6} \\
 & \Rightarrow P\left( {{E}_{3}}|X \right)=\dfrac{0.42}{0.1\times 0.2+0.2\times 0.4+0.7\times 0.6}=\dfrac{0.42}{0.52}=\dfrac{21}{26} \\
\end{align}\]
So the correct option is D.

Note: We note that the events ${{E}_{1}},{{E}_{2}},{{E}_{3}},...,{{E}_{n}}\left( {{E}_{i}}\ne \Phi \right)$ have to be mutually exclusive (do not occur at the same time) and exhaustive (sum of events sets is the sample space) for us to use Bayes’ theorem. . We see in this problem that she can only study in one day for a certain hours. So the events ${{E}_{1}},{{E}_{2}},{{E}_{3}}$ are mutually exclusive. We also see that $P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)+P\left( {{E}_{3}} \right)=0.1+0.2+0.7=1$ . So the events are exhaustive too. We also note that the new event must not be improbable which means $P\left( X \right)\ne 0$ .