# 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.

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.

Last updated date: 29th Sep 2023

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