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Probability question

2011 · Shift 2 · Q29
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  5. /2011 · Shift 2 · Q29

Probability question

2011 · Shift 2 · Q29

JEE AdvancedMathematicsProbabilityMultiple correct+4 / −1
Let EEE and FFF be two independent events. The probability that exactly one of them occurs is  1125\,{{11} \over {25}}2511​ and the probability of none of them occurring is  225\,{{2} \over {25}}252​. If P(T)P(T)P(T) denotes the probability of occurrence of the event T,T,T, then
  1. A
    P(E)=45,P(F)=35P\left( E \right) = {4 \over 5},P\left( F \right) = {3 \over 5}P(E)=54​,P(F)=53​
  2. B
    P(E)=15,P(F)=25P\left( E \right) = {1 \over 5},P\left( F \right) = {2 \over 5}P(E)=51​,P(F)=52​
  3. C
    P(E)=25,P(F)=15P\left( E \right) = {2 \over 5},P\left( F \right) = {1 \over 5}P(E)=52​,P(F)=51​
  4. D
    P(E)=35,P(F)=45P\left( E \right) = {3 \over 5},P\left( F \right) = {4 \over 5}P(E)=53​,P(F)=54​
View written solutionFree

Correct answer: A, D

  1. Let P(E)=p,P(F)=q.P(E)=p, \quad P(F)=q.P(E)=p,P(F)=q. Since EEE and FFF are independent, P(E∩F)=pq.P(E\cap F)=pq.P(E∩F)=pq.

  2. Probability that none occurs is given as P(E′∩F′)=225.P(E'\cap F')=\frac{2}{25}.P(E′∩F′)=252​. Using independence, P(E′∩F′)=(1−p)(1−q)=225.P(E'\cap F')=(1-p)(1-q)=\frac{2}{25}. P(E′∩F′)=(1−p)(1−q)=252​.

  3. Probability that exactly one occurs is P(E∩F′)+P(E′∩F).P(E\cap F')+P(E'\cap F).P(E∩F′)+P(E′∩F). Since the events are independent, p(1−q)+(1−p)q=1125.p(1-q)+(1-p)q=\frac{11}{25}. p(1−q)+(1−p)q=2511​. Simplifying, p+q−2pq=1125.p+q-2pq=\frac{11}{25}. p+q−2pq=2511​.

  4. Also, from step 2, 1−p−q+pq=225.1-p-q+pq=\frac{2}{25}.1−p−q+pq=252​. So, p+q−pq=1−225=2325.p+q-pq=1-\frac{2}{25}=\frac{23}{25}. p+q−pq=1−252​=2523​.

  5. Now subtract the equation p+q−2pq=1125p+q-2pq=\frac{11}{25}p+q−2pq=2511​ from p+q−pq=2325:p+q-pq=\frac{23}{25}:p+q−pq=2523​: pq=2325−1125=1225.pq=\frac{23}{25}-\frac{11}{25}=\frac{12}{25}. pq=2523​−2511​=2512​.

  6. Then p+q=2325+1225=3525=75.p+q=\frac{23}{25}+\frac{12}{25}=\frac{35}{25}=\frac{7}{5}. p+q=2523​+2512​=2535​=57​.

  7. So ppp and qqq satisfy x2−(p+q)x+pq=0x^2-(p+q)x+pq=0x2−(p+q)x+pq=0 i.e. x2−75x+1225=0.x^2-\frac{7}{5}x+\frac{12}{25}=0. x2−57​x+2512​=0. Multiply by 252525: 25x2−35x+12=0.25x^2-35x+12=0.25x2−35x+12=0.

  8. Factorizing, 25x2−35x+12=(5x−3)(5x−4)=0.25x^2-35x+12=(5x-3)(5x-4)=0.25x2−35x+12=(5x−3)(5x−4)=0. Hence, x=35orx=45.x=\frac35 \quad \text{or} \quad x=\frac45. x=53​orx=54​.

    Therefore, {P(E),P(F)}={35,45}.\{P(E),P(F)\}=\left\{\frac35,\frac45\right\}. {P(E),P(F)}={53​,54​}.

  9. Check options:

    • A: P(E)=45,  P(F)=35P(E)=\frac45,\; P(F)=\frac35P(E)=54​,P(F)=53​ ✓
    • B: P(E)=15,  P(F)=25P(E)=\frac15,\; P(F)=\frac25P(E)=51​,P(F)=52​ ✗
    • C: P(E)=25,  P(F)=15P(E)=\frac25,\; P(F)=\frac15P(E)=52​,P(F)=51​ ✗
    • D: P(E)=35,  P(F)=45P(E)=\frac35,\; P(F)=\frac45P(E)=53​,P(F)=54​ ✓

Therefore, the correct options are A and D.

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