
The probability that a student selected at random from a class will pass in Mathematics is 4/5, and the probability that he/she passes in Mathematics and Computer Science is 1/2. What is the probability that he/she will pass in Computer Science if it is known that he has passed in Mathematics?
Answer
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Hint: In this question, we need to determine the probability that he/she will pass in Computer Science if it is known that he has passed in Mathematics such that the probability that a student selected at random from a class will pass in Mathematics is 4/5, and the probability that he/she passes in Mathematics and Computer Science is 1/2. For this, we will use the concept of conditional probability such that one event has been confirmed here.
Complete step-by-step answer:
Let A be the event that the candidate passes in Mathematics.
Let B be the event that the candidate passes in Computer Science.
According to the question, the probability that the candidate passes in mathematics is 4/5 so, we can write, $ P(A) = \dfrac{4}{5} $
Also, it has been given that the probability that the candidate passes in Mathematics and Computer Science is 1/2. So, we can write, $ P(A \cap B) = \dfrac{1}{2} $ .
Now, as per the demand of the question we need to calculate the probability that the candidate will pass in Computer Science if it is known that he has passed in Mathematics so, using conditional probability we can write
$ P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) $ where, $ P\left( {{\text{B/A}}} \right) $ denotes that the event A has already been confirmed.
Substituting the values in the equation $ P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) $ , we get
$
P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) \\
\Rightarrow \dfrac{1}{2} = P\left( {{\text{B/A}}} \right) \times \dfrac{4}{5} \\
\Rightarrow P\left( {{\text{B/A}}} \right) = \dfrac{{\left( {\dfrac{4}{5}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} \\
\Rightarrow P\left( {{\text{B/A}}} \right) = \dfrac{5}{8} \;
$
Hence, the probability that he/she will pass in Computer Science if it is known that he has passed in Mathematics is $ \dfrac{5}{8} $ .
So, the correct answer is “ $ \dfrac{5}{8} $ ”.
Note: Conditional probability is the measure of the probability of an event considering that another event has already occurred. Here, two events are not dependent on each other but the series of occurrences of events is one after the other which means that the succeeding events will occur only if the preceding events have taken place.
Complete step-by-step answer:
Let A be the event that the candidate passes in Mathematics.
Let B be the event that the candidate passes in Computer Science.
According to the question, the probability that the candidate passes in mathematics is 4/5 so, we can write, $ P(A) = \dfrac{4}{5} $
Also, it has been given that the probability that the candidate passes in Mathematics and Computer Science is 1/2. So, we can write, $ P(A \cap B) = \dfrac{1}{2} $ .
Now, as per the demand of the question we need to calculate the probability that the candidate will pass in Computer Science if it is known that he has passed in Mathematics so, using conditional probability we can write
$ P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) $ where, $ P\left( {{\text{B/A}}} \right) $ denotes that the event A has already been confirmed.
Substituting the values in the equation $ P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) $ , we get
$
P(A \cap B) = P\left( {{\text{B/A}}} \right) \times P(A) \\
\Rightarrow \dfrac{1}{2} = P\left( {{\text{B/A}}} \right) \times \dfrac{4}{5} \\
\Rightarrow P\left( {{\text{B/A}}} \right) = \dfrac{{\left( {\dfrac{4}{5}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} \\
\Rightarrow P\left( {{\text{B/A}}} \right) = \dfrac{5}{8} \;
$
Hence, the probability that he/she will pass in Computer Science if it is known that he has passed in Mathematics is $ \dfrac{5}{8} $ .
So, the correct answer is “ $ \dfrac{5}{8} $ ”.
Note: Conditional probability is the measure of the probability of an event considering that another event has already occurred. Here, two events are not dependent on each other but the series of occurrences of events is one after the other which means that the succeeding events will occur only if the preceding events have taken place.
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