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In a college, 70 % students pass in physics, 75 % pass in mathematics and 10 % students fail in both. One student is chosen at random. What is the probability that the student passes in mathematics given that he passes in physics?

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Hint: Let us denote P be the event that the student passes in physics and let us denote the M be the event that the student passes in mathematics. The probability that the student passes in mathematics given that he passes in physics is given by the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$. Using this formula, we can solve this question.

Complete step by step solution:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
If we are given two events A and B, then the probability of event A given that event B will also occur is given by the formula,
$P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ . . . . . . . . . . . . . (1)
Also, in probability, we have a formula $P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cup B \right)$ . . . . . . . . . (2)
For this question, let us denote P be the event that the student passes in physics and let us denote the M be the event that the student passes in mathematics.
It is given that 70 % students pass in physics, so, $P\left( P \right)=\dfrac{70}{100}$.
$\Rightarrow P\left( M \right)=\dfrac{7}{10}$
It is given that 75 % students pass in mathematics, so, $P\left( M \right)=\dfrac{75}{100}$.
$\Rightarrow P\left( P \right)=\dfrac{3}{4}$
Also, it is given that 10 % of the students fail in both subjects. So, we can say 90% of the students passed either in physics, or in mathematics, or in both.
$\begin{align}
  & \Rightarrow P\left( M\cup P \right)=\dfrac{90}{100} \\
 & \Rightarrow P\left( M\cup P \right)=\dfrac{9}{10} \\
\end{align}$
Using formula (2), we get,
\[\begin{align}
  & P\left( M\cap P \right)=\dfrac{7}{10}+\dfrac{3}{4}-\dfrac{9}{10} \\
 & \Rightarrow P\left( M\cap P \right)=\dfrac{28+30-36}{40} \\
 & \Rightarrow P\left( M\cap P \right)=\dfrac{22}{40} \\
 & \Rightarrow P\left( M\cap P \right)=\dfrac{11}{20} \\
\end{align}\]
Using formula (1), the probability that the student passes in mathematics given that he passes in physics is equal to,
$\begin{align}
  & P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)} \\
 & \Rightarrow P\left( M|P \right)=\dfrac{\dfrac{11}{20}}{\dfrac{7}{10}} \\
 & \Rightarrow P\left( M|P \right)=\dfrac{11}{14} \\
\end{align}$
Hence, the answer is $\dfrac{11}{14}$.

Note: There is a possibility that one may commit a mistake while using the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$. Instead of the correct formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$, it is possible that one may use the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( M \right)}$ which will lead us to an incorrect answer.

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