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A student is given with a true-false exam with 10 questions. If he gets 8 or more correct answers , he passes the exam. Given that he guesses the answer of each question, the probability that he passes the exam is $\dfrac{k}{128}$. Find the value of k.

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: First, we need to find the total number of ways by which he passes the exam. Then, we need to find the cases for which he gets a correct of at least 8 answers which gives us the favourable cases. Then, we need to use the formula of combination as $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ to get the favourable cases value. Then, to get the probability that he guesses the answer of the questions at least 8 times correct out of total cases is given by the ratio of the total number of the favourable cases to the total number of cases.

Complete step-by-step answer:
In this question, we are supposed to find the probability of the situation that he passes the exam.
Moreover, this probability is given in the form of k as $\dfrac{k}{128}$and we need to find out the value of k.
Now, we need to find the total number of ways by which he passes the exam.
So, we know that there are only two possibilities of the answer that is true or false.
Then, the total number of ways is given by:
${{2}^{10}}=1024$
Then, we need to find the cases for which he gets a correct of at least 8 answers which gives us the favourable cases.
So, to find it we use the concept of combination as:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$
Now, the favourable cases are with values r=8, 9, 10 and n=10 as:
${}^{10}{{C}_{8}}+{}^{10}{{C}_{9}}+{}^{10}{{C}_{10}}$
So, solve the above expression to get the total number of the favourable cases as:
$\begin{align}
  & \dfrac{10!}{\left( 10-8 \right)!8!}+\dfrac{10!}{\left( 10-9 \right)!9!}+\dfrac{10!}{\left( 10-10 \right)!10!} \\
 & \Rightarrow \dfrac{10\times 9\times 8!}{2!8!}+\dfrac{10\times 9!}{1!9!}+\dfrac{10!}{0!10!} \\
 & \Rightarrow \dfrac{10\times 9}{2}+\dfrac{10}{1}+1 \\
 & \Rightarrow 45+10+1 \\
 & \Rightarrow 56 \\
\end{align}$
Now, to get the probability that he guesses the answer of the questions at least 8 times correct out of total cases is given by the ratio of the total number of the favourable cases to the total number of cases:
$\begin{align}
  & P=\dfrac{56}{1024} \\
 & \Rightarrow P=\dfrac{7}{128} \\
\end{align}$
Now, compare the above calculated value with the $\dfrac{k}{128}$.
$\dfrac{k}{128}=\dfrac{7}{128}$
So, by comparing the above equation, we get
$k=7$
Hence, the value of k is 7.


Note: Here, we need to keep in mind the fact given in the question as the condition says that student needs to answer 8 or more correct answers to get the probability. So, there is another approach to get the final answer as we can subtract the probability of answering at most 7 correctly from 1 as 1 is always the total probability. So, the alternative approach is as:
$1-\left( {}^{10}{{C}_{0}}+{}^{10}{{C}_{1}}+{}^{10}{{C}_{2}}+{}^{10}{{C}_{3}}+{}^{10}{{C}_{4}}+{}^{10}{{C}_{5}}+{}^{10}{{C}_{6}}+{}^{10}{{C}_{7}} \right)$ which also gives the same answer.