Question

guesses. Let
be the probability
In answering a question on a multiple choice
test, a student either knows the answer or
3
be the probability that he
5
2
knows the answer and
5
that he guesses. Assuming that probability of
student who guessing the correct answer is
1
what is the probability that the student
3
knows the answer given that he answered it
correctly?
[CBSE 2015]
9​

Answer 1

Appropriate Question:

In answering a question on a multiple choice test, a student either knows the answer or guess the answer. If 3/5 be the probability that he guesses. and 2/5 be the probability knows the answer. Assuming that probability of student who guessing the correct answer is 1/3, what is the probability that the student knows the answer given that he answered it correctly?

Answer:

The probability that the student knows the answer given that he answered it correctly is $\dfrac{2}{3}$.

Step-by-step explanation:

Let assume the following events

E₁ : Student knows the answer to a question.

E₂ : Student guess the answer to a question.

E : He answered the question correctly.

Now,

$\sf \: P(E_1) = \dfrac{2}{5}$

$\sf \: P(E_2) =  \dfrac{3}{5}$

Now, Further given that

$\sf \: P(E  \: \mid \: E_1) = 1$

$\sf \: P(E  \: \mid \: E_2) = \dfrac{1}{3}$

Now, By definition of Bayes Theorem, we have

$\sf P(E_1\mid \: E) = \dfrac{P(E_1) . P(E  \mid E_1)}{P(E_1) . P(E  \mid E_1) + P(E_2) . P(E  \mid E_2)}$

So, on substituting the values, we get

$\sf P(E_1 \mid \: E) = \dfrac{\dfrac{2}{5}  \times 1}{\dfrac{2}{5}  \times 1  + \dfrac{3}{5}  \times \dfrac{1}{3}}$

$\sf P(E_1 \mid \: E) = \dfrac{2}{2 + 1}$

$\sf\implies \sf P(E_1 \mid \: E) = \dfrac{2}{3}$

Thus,

$\sf\implies \sf Required\:probability\:is\:\dfrac{2}{3}$

Hence,

The probability that the student knows the answer given that he answered it correctly is $\dfrac{2}{3}$.

Answer 2

Question:-

In answering a question on a multiple choice test, a student either knows the answer or guess the answer. If 3/5 be the probability that he guesses. and 2/5 be the probability knows the answer. Assuming that probability of student who guessing the correct answer is 1/3, what is the probability that the student knows the answer given that he answered it correctly?

Correct answer:-

The probability that the student knows the answer given that he answered it correctly is \dfrac{2}{3}

3

2

.

Step-by-step explanation:

Let assume the following events

E₁ : Student knows the answer to a question.

E₂ : Student guess the answer to a question.

E : He answered the question correctly.

Now,

\begin{gathered}\sf \: P(E_1) = \dfrac{2}{5} \\ \\ \end{gathered}

P(E

1

)=

5

2

\begin{gathered}\sf \: P(E_2) = \dfrac{3}{5} \\ \\ \end{gathered}

P(E

2

)=

5

3

Now, Further given that

\begin{gathered}\sf \: P(E \: \mid \: E_1) = 1 \\ \\ \end{gathered}

P(E ∣E

1

)=1

\begin{gathered}\sf \: P(E \: \mid \: E_2) = \dfrac{1}{3} \\ \\ \end{gathered}

P(E ∣E

2

)=

3

1

Now, By definition of Bayes Theorem, we have

\begin{gathered}\sf P(E_1\mid \: E) = \dfrac{P(E_1) . P(E \mid E_1)}{P(E_1) . P(E \mid E_1) + P(E_2) . P(E \mid E_2)} \\ \\ \end{gathered}

P(E

1

∣E)=

P(E

1

).P(E ∣E

1

)+P(E

2

).P(E ∣E

2

)

P(E

1

).P(E ∣E

1

)

So, on substituting the values, we get

\begin{gathered}\sf P(E_1 \mid \: E) = \dfrac{\dfrac{2}{5} \times 1}{\dfrac{2}{5} \times 1 + \dfrac{3}{5} \times \dfrac{1}{3}} \\ \\ \end{gathered}

P(E

1

∣E)=

5

2

×1 +

5

3

×

3

1

5

2

×1

\begin{gathered}\sf P(E_1 \mid \: E) = \dfrac{2}{2 + 1} \\ \\ \end{gathered}

P(E

1

∣E)=

2+1

2

\begin{gathered}\sf\implies \sf P(E_1 \mid \: E) = \dfrac{2}{3} \\ \\ \end{gathered}

⟹P(E

1

∣E)=

3

2

Thus,

\begin{gathered}\sf\implies \sf Required\:probability\:is\:\dfrac{2}{3} \\ \\ \end{gathered}

⟹Requiredprobabilityis

3

2

Hence,

The probability that the student knows the answer given that he answered it correctly is \dfrac{2}{3}

3

2

.

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