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Probability question

2017 · 9 Apr · Shift 1 · Q33
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Probability question

2017 · 9 Apr · Shift 1 · Q33

JEE MainMathematicsProbabilityMCQ+4 / −1
Let E and F be two independent events. The probability that both E and F happen is 112{1 \over {12}}121​ and the probability that neither E nor F happens is 12{1 \over {2}}21​, then a value of P(E)P(F){{P\left( E \right)} \over {P\left( F \right)}}P(F)P(E)​ is :
  1. A
    43{4 \over 3}34​
  2. B
    32{3 \over 2}23​
  3. C
    13{1 \over 3}31​
  4. D
    512{5 \over 12}125​
View written solutionFree

Correct answer: A

  1. Let P(E)=p,P(F)=qP(E)=p, \quad P(F)=qP(E)=p,P(F)=q Since EEE and FFF are independent, P(E∩F)=pq=112P(E\cap F)=pq=\frac{1}{12}P(E∩F)=pq=121​

  2. The probability that neither EEE nor FFF happens is P(E′∩F′)=12P(E'^{}\cap F'{})=\frac{1}{2}P(E′∩F′)=21​ For independent events, complements are also independent, so P(E′∩F′)=(1−p)(1−q)=12P(E'^{}\cap F'{})=(1-p)(1-q)=\frac{1}{2}P(E′∩F′)=(1−p)(1−q)=21​

  3. Expand: 1−p−q+pq=121-p-q+pq=\frac{1}{2}1−p−q+pq=21​ Using pq=112pq=\frac{1}{12}pq=121​, 1−p−q+112=121-p-q+\frac{1}{12}=\frac{1}{2}1−p−q+121​=21​ 1312−p−q=12\frac{13}{12}-p-q=\frac{1}{2}1213​−p−q=21​ p+q=1312−12=1312−612=712p+q=\frac{13}{12}-\frac{1}{2}=\frac{13}{12}-\frac{6}{12}=\frac{7}{12}p+q=1213​−21​=1213​−126​=127​

  4. Now ppp and qqq satisfy p+q=712,pq=112p+q=\frac{7}{12}, \quad pq=\frac{1}{12}p+q=127​,pq=121​ So they are roots of x2−712x+112=0x^2-\frac{7}{12}x+\frac{1}{12}=0x2−127​x+121​=0 Multiply by 121212: 12x2−7x+1=012x^2-7x+1=012x2−7x+1=0 Factor: (3x−1)(4x−1)=0(3x-1)(4x-1)=0(3x−1)(4x−1)=0 Hence, x=13orx=14x=\frac{1}{3} \quad \text{or} \quad x=\frac{1}{4}x=31​orx=41​ Therefore, {p,q}={13,14}\{p,q\}=\left\{\frac{1}{3},\frac{1}{4}\right\}{p,q}={31​,41​}

  5. Thus, P(E)P(F)=pq\frac{P(E)}{P(F)}=\frac{p}{q}P(F)P(E)​=qp​ can be either 1/31/4=43\frac{1/3}{1/4}=\frac{4}{3}1/41/3​=34​ or 1/41/3=34\frac{1/4}{1/3}=\frac{3}{4}1/31/4​=43​

  6. Among the given options, only 43\frac{4}{3}34​ appears.

Therefore, the correct option is A.

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