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  • A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4, respectively. If the probability of their making a common error is, 1/20 and they obtain the same answer, then the probability of their answer to be correct is A. 1/12

A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4, respectively. If the probability of their making a common error is, 1/20 and they obtain the same answer, then the probability of their answer to be correct is
A.\frac{1}{12}

B.\frac{1}{40}

C.\frac{13}{120}

D.\frac{10}{13}

Answers (2)

Solution

Let E be the event that student ‘A’ solves the problem correctly.

∴ P(E) = 1/3
In the same way if we denote the event of ’B’ solving the problem correctly with F

We get  P(F) = 1/4
Since both the events are independent.
∴ Probability that both the students solve the question correctly can be represented as-

P(E \cap F)=\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}=P\left(E_{1}\right)\(say) \ \\\(we \ can \ multiply \ because \ events \ are \ independent \)
\therefore Probability that both the students could not solve the question correctly can be represented as-
\mathrm{P}\left(\mathrm{E}^{\prime} \cap \mathrm{F}^{\prime}\right)=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}=\mathrm{P}\left(\mathrm{E}_{2}\right)\{\mathrm{say}\}
Given: probability of making a common error and both getting same answer.
If they are making an error, we can be sure that answer coming out is wrong.
Let S be the event of getting same answer.
\therefore  above situation can be represented using conditional probability.
\mathrm{P}\left(\mathrm{S} \mid \mathrm{E}_{2}\right)=1 / 20
And if their answer is correct obviously, they will get same answer.
\therefore P\left(S \mid E_{1}\right)=1

To find- the probability of getting a correct answer if they committed a common error and got the same answer.
By observing our requirement and availability of equations, we can use Bayes theorem to solve this.
\therefore   Using Bayes theorem, we get-
P\left(E_{1} \mid S\right)=\frac{P\left(E_{1}\right) P\left(S \mid E_{1}\right)}{P\left(E_{1}\right) P\left(S \mid E_{1}\right)+P\left(E_{2}\right) P\left(S \mid E_{2}\right)}
Substituting the values from above -
P\left(E_{1} \mid S\right)=\frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1+\frac{1}{2} \times \frac{1}{20}}=\frac{\frac{1}{12}}{\frac{1}{12}+\frac{1}{40}}=\frac{10}{13}

Our answers clearly match with option D.
∴ Option (D) is the only correct choice.

Posted by

infoexpert22

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Let E1 be the event that both A and B solve correctly.

P(E1)=1/3\times1/4=1/12

Let E2 be the event when both  A and B solve incorrectly.

P(E2)=2/3\times3/4=1/2

Let F be the prpbability that both A and B got the correct answer.

P(F/E1)=1                P(F/E2)=1/20

Probability of getting correct answer by both A and B given that they both got the same answer.

P(E1/F)= (1\times1/12)/[(1\times1/12)+(1/2\times1/20)]

P(E1/F)=(1/12)/[(1/12)+(1/40)]

P(E1/F)=(1/12)/[(10/120)+(3/120)]

P(E1/F)=(1/12)/(13/120)

P(E1/F)=(1/12)\times(120/13)

P(E1/F)=10/13

Therefore the Probability of getting a correct answer by both A and B given that they both got the same answer is 10/13.

 

 

 

Posted by

Atchaya R

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