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Question: For a student to be qualified, he must pass at least two out of three exams. The probability that he...

For a student to be qualified, he must pass at least two out of three exams. The probability that he will pass the 1st exam is p. If he fails in one of the exams then the probability of his passing in the next exam is p2\dfrac{p}{2} otherwise it will remain the same. Find the probability that he will qualify.

Explanation

Solution

Let,
E1={E_1} = Event that the students will pass the math exam. (Where a=1,2,3a = 1,2,3 )
E2={E_2} = Event that the students will qualify
Then a student can qualify in anyone of the following ways:

  1. He passed the first exam and second exam.
  2. He passes the first exam, fails in the second exam but passes in the third exam.
  3. He fails the first exam, passes in the second exam and third exam.
    Then using formula, P(E)=[P(E1)×P(E2E1)]+[P(E1)×P(E2′E1)×P(E3E2′)]+[P(E1′)×P(E2E1′)×P(E3E2)]P\left( E \right) = \left[ {P\left( {{E_1}} \right) \times P\left( {\dfrac{{{E_2}}}{{{E_1}}}} \right)} \right] + \left[ {P\left( {{E_1}} \right) \times P\left( {\dfrac{{{E_2}'}}{{{E_1}}}} \right) \times P\left( {\dfrac{{{E_3}}}{{{E_2}'}}} \right)} \right] + \left[ {P\left( {{E_1}'} \right) \times P\left( {\dfrac{{{E_2}}}{{{E_1}'}}} \right) \times P\left( {\dfrac{{{E_3}}}{{{E_2}}}} \right)} \right] find the probability that the student will qualify.

Complete step by step solution:
Here, we have given that a student to be qualified, must pass at least two out of three exams.
The probability of passing the 1st exam is p.
∵\because The student fails in one of the exams then the probability of passing in the next exam is p2\dfrac{p}{2} otherwise it will remain the same.
Then, we have to find the probability that the student will qualify.
Now, let us consider
E1={E_1} = Event that the students will pass the math exam. (Where a=1,2,3a = 1,2,3 )
E2={E_2} = Event that the students will qualify
Then a student can qualify in anyone of the following ways:

  1. He passed the first exam and second exam.
  2. He passes the first exam, fails in the second exam but passes in the third exam.
  3. He fails the first exam, passes in the second exam and third exam.
    ⇒P(E)=[P(E1)×P(E2E1)]+[P(E1)×P(E2′E1)×P(E3E2′)]+[P(E1′)×P(E2E1′)×P(E3E2)]\Rightarrow P\left( E \right) = \left[ {P\left( {{E_1}} \right) \times P\left( {\dfrac{{{E_2}}}{{{E_1}}}} \right)} \right] + \left[ {P\left( {{E_1}} \right) \times P\left( {\dfrac{{{E_2}'}}{{{E_1}}}} \right) \times P\left( {\dfrac{{{E_3}}}{{{E_2}'}}} \right)} \right] + \left[ {P\left( {{E_1}'} \right) \times P\left( {\dfrac{{{E_2}}}{{{E_1}'}}} \right) \times P\left( {\dfrac{{{E_3}}}{{{E_2}}}} \right)} \right]
    ⇒P(E)=p2+p(1−p).p2+(1−p).p2.p\Rightarrow P\left( E \right) = {p^2} + p\left( {1 - p} \right).\dfrac{p}{2} + \left( {1 - p} \right).\dfrac{p}{2}.p
    ⇒P(E)=2p2+p2−p3+p2−p32\Rightarrow P\left( E \right) = \dfrac{{2{p^2} + {p^2} - {p^3} + {p^2} - {p^3}}}{2}
    ⇒P(E)=2p2−p3\Rightarrow P\left( E \right) = 2{p^2} - {p^3}

∴\therefore The probability that the student will qualify is 2p2−p32{p^2} - {p^3}.

Note:
Alternate Method:
let us consider
E1={E_1} = Event that the students will pass the ath{a^{th}}exam. (Where a=1,2,3a = 1,2,3 )
E2={E_2} = Event that the students will qualify
Then a student can qualify in anyone of the following ways:

  1. He passed the first exam and second exam.
  2. He passes the first exam, fails in the second exam but passes in the third exam.
  3. He fails the first exam, passes in the second exam and third exam.
    The probability that the student pass all the three exams: PPP
    The probability that the student fails in one of the exams: PPF, PFP, FPP
    ⇒P(E)=PPP+PPF+PFP+FPP\Rightarrow P\left( E \right) = PPP + PPF + PFP + FPP
    ⇒P(E)=(PPP)+(PP(1−p))+(P(1−p)p2)+((1−p)p2p)\Rightarrow P\left( E \right) = \left( {PPP} \right) + \left( {PP\left( {1 - p} \right)} \right) + \left( {P\left( {1 - p} \right)\dfrac{p}{2}} \right) + \left( {\left( {1 - p} \right)\dfrac{p}{2}p} \right)
    ⇒P(E)=P3+P2−P3+P22−P32+P22−P32\Rightarrow P\left( E \right) = {P^3} + {P^2} - {P^3} + \dfrac{{{P^2}}}{2} - \dfrac{{{P^3}}}{2} + \dfrac{{{P^2}}}{2} - \dfrac{{{P^3}}}{2}
    ⇒P(E)=2P3−P2\Rightarrow P\left( E \right) = 2{P^3} - {P^2}