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In a maths paper there are 3 sections A, B & C. Section A is compulsory. Out of sections B & C a student has to attempt any one. Passing in the paper means passing in A & passing in B or C. The probability of the student passing in A, B & C are p, q & $\dfrac{1}{2}$ respectively. If the probability that the student is successful is $\dfrac{1}{2}$ then, which of the following is false?
(This question has multiple correct options)
A) $p = q = 1$
B) $p = q = \dfrac{1}{2}$
C) $p = 1,q = 0$
D) \[p = 1,q = \dfrac{1}{2}\]

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Last updated date: 21st Jun 2024
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Answer
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Hint: We will first of all list down both the cases when a student will be termed as “pass”. After that, we will find the probability of both the cases to occur. Now, on adding both the probabilities, we must get $\dfrac{1}{2}$.

Complete step-by-step solution:
We are given that: “Passing in the paper means passing in A & passing in B or C”.
Therefore, the cases when a students will be successful is:
Case 1: When he / she passes in section A and section B
Case 2: When he / she passes in section A and section C.
We are also given that: the probability of the student passing in A, B & C are p, q & $\dfrac{1}{2}$ respectively.
$\therefore $ Probability of case 1 to occur will be = $p \times q = pq$
And, Probability of case 2 to occur will be = $p \times 0.5 = 0.5p$
Now, Total probability of the student being successful is the sum of the probability of case 1 and case 2.
$\therefore \Pr obability = pq + 0.5p$
We are already given that the probability that the student is successful is $\dfrac{1}{2}$.
$\therefore pq + 0.5p = \dfrac{1}{2}$
$ \Rightarrow p\left( {q + \dfrac{1}{2}} \right) = \dfrac{1}{2}$
Now, we will check all the options one by one.
Option A: If $p = q = 1$, then $p\left( {q + \dfrac{1}{2}} \right) = 1\left( {\dfrac{3}{2}} \right) = \dfrac{3}{2}$ which is not equal to RHS. Hence, A is false.
Option B: If $p = q = \dfrac{1}{2}$, then $\dfrac{1}{2}\left( {\dfrac{1}{2} + \dfrac{1}{2}} \right) = \dfrac{1}{2}\left( 1 \right) = \dfrac{1}{2}$ which is equal to RHS. Hence, B is true.
Option C: If $p = 1,q = 0$, then $1\left( {0 + \dfrac{1}{2}} \right) = 1\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}$ which is equal to RHS. Hence, C is true.
Option D: If \[p = 1,q = \dfrac{1}{2}\], then $1\left( {\dfrac{1}{2} + \dfrac{1}{2}} \right) = 1\left( 1 \right) = 1$ which is not equal to RHS. Hence, D is false.

$\therefore $ The correct options are A and D.

Note: The students must keep in mind that we cannot firmly find the values of p and q, we will have to check the values only by putting in the options and whichever would not satisfy the equation we found will be the answers.
Since, this is a multiple correct question, students must not stop after checking the first part only and should check each and every part thoroughly.