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Question: Of the 25 questions in a unit, a student has worked out only 20. In a sessional test of that unit, t...

Of the 25 questions in a unit, a student has worked out only 20. In a sessional test of that unit, two questions were asked by the teacher. The probability that the student can solve both the question correctly, is
A. 825\dfrac{8}{{25}}
B. 1725\dfrac{{17}}{{25}}
C. 910\dfrac{9}{{10}}
D. 1930\dfrac{{19}}{{30}}

Explanation

Solution

Here, we will first find the number of choices the teacher has out of 25 questions to ask the student. Then we will find the total number of questions a student could have answered if two questions were asked using this, we will find the probability that the student answered correctly in the sessional test.

Complete Step by Step Solution:
Here we need to find the probability that the student could correct answer the question.
it is given that:
Total number of questions in a unit =25 = 25
Total number of questions the students could answer =20 = 20
Now, we will find the number of choices the teacher has out of 25 questions to ask the student and that will be the total number of outcomes.
Number of choices the teacher has to ask the student =25C2 = {}^{25}{C_2}
Now using the formula of combination nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we get
Therefore, we get
\Rightarrow Number of choices the teacher has to ask the student =25!2!(252)! = \dfrac{{25!}}{{2!\left( {25 - 2} \right)!}}
On further simplifying the terms, we get
\Rightarrow Number of choices the teacher has to ask the student =25!2!×23! = \dfrac{{25!}}{{2! \times 23!}}
Computing the factorials, we get
\Rightarrow Number of choices the teacher has to ask the student =25×24×23!2×1×23! = \dfrac{{25 \times 24 \times 23!}}{{2 \times 1 \times 23!}}
On further simplification, we get
$$$$ Number of choices the teacher has to ask the student =25×12 = 25 \times 12
On multiplying the numbers, we get
\Rightarrow Number of choices the teacher has to ask the student =300 = 300
Total number of questions student could have answered =20C2 = {}^{20}{C_2}
Now using the formula of combination nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we get
\Rightarrow Total number of questions that a student could have answered =20!2!(202)! = \dfrac{{20!}}{{2!\left( {20 - 2} \right)!}}
On further simplifying the terms, we get
\Rightarrow Total number of questions that a student could have answered =20!2!×18! = \dfrac{{20!}}{{2! \times 18!}}
Computing the factorial, we get
\Rightarrow Total number of questions that a student could have answered =20×19×18!2×1×18! = \dfrac{{20 \times 19 \times 18!}}{{2 \times 1 \times 18!}}
On further simplification, we get
\Rightarrow Total number of questions that a student could have answered =10×19 = 10 \times 19
On multiplying the numbers, we get
\Rightarrow Total number of questions that a student could have answered =190 = 190
Now, we will find the probability that the student answered correctly in the test which will be equal to the ratio of the number of choices the teacher has out of 25 questions to ask the student to the total number of questions the student could have answered if two questions were asked.
Therefore,
Probability that the student answered correctly =190300 = \dfrac{{190}}{{300}}
On further simplification, we get
\Rightarrow Probability that the student answered correctly =1930 = \dfrac{{19}}{{30}}

Hence, the correct option is option D.

Note:
To solve this problem, we need to know about probability and its properties. Probability is defined as the ratio of the required number of outcomes to the total number of outcomes. We need to keep in mind that the value of probability cannot be greater than 1 and the value of probability cannot be negative. In addition, the probability of a sure event is always one and the event with a zero probability never occurs.