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Grade 10Algebra

In a test an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he make a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it, is 1/8. Find the probability that he knew the answer to the question given that he correctly answered it.

Hrishant Goswami

12 Years agoGrade 10

1 Answer

Jitender Pal

12 Years ago

Hello Student,

Please find the answer to your question

KEY CONCEPT :

Baye’s theorem: E₁, E₂, E₃, . . . . . . . . . . . . . . ,E_n are mutually exclusive and exhaustive events and E is an event which takes place in conjunction with any one of E₁ then the probability of the event E₁ happening when the event E has taken place is given by

P (E₁| E) = $\frac{P(E_{1})P(E|E_{1})}{\sum_{i=1}^{n}P(E_1)P(E|E_1)}$

Let us define the events :

A₁ ≡ the examinee guesses the answer,

A₂ ≡ the examinee copies the answer

A₃ ≡ the examinee knows the answer,

A ≡ the examinee answers correctly

ATQ, P (A₁) = 1/3; P (A₂) = 1/6

As any one happens out of A₁, A₂, A₃, these are mutually exclusive and exhaustive events.

∴ P (A₁) + P (A₂) + P (A₃) = 1

⇒ P (A₃) = 1 – 1/3 – 1/6 = 6 - 2 – 1/6 = 3/6 = 1/2

Also we have, P(A| A₁) = 1/4

[∵ out of 4 choices only one is correct.] P (A| A₂) = 1/8

(given) P (A| A₃) = 1

[If examinee knows the ans., it is correct. i.e. true event]

To find P (A₃| A). By Baye’s thm. P (A₃| A)

= P (A₃| A) P (A₃)/P (A| A₁) P(A₁) + P (A| A₂) P(A₂) + P (A| A₃) P(A₃)

= $\frac{1\frac{1}{2}}{\frac{1}{4}.\frac{1}{3}+\frac{1}{8}.\frac{1}{6}+\frac{1.1}{2}}$ = 1/2/29/48 = 1/2 x 18/29 = 24/29.

Thanks
Jitender Pal
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