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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

JEE Advanced 2011 · Paper 2 · Mathematics · Probability

Let E and F be two independent events. The probability that exactly...

Let EE and FF be two independent events. The probability that exactly one of them occurs is 11/2511 / 25 and the probability of none of them occurring is 2/252 / 25. If P(T)P(T) denotes the probability of occurrence of the event TT, thenLet E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probabilityLet E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probabilityLet E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probabilityLet E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probability

Options

  1. AP(E)=45,P(F)=35P(E)=\frac{4}{5}, P(F)=\frac{3}{5}
  2. BP(E)=15,P(F)=25P(E)=\frac{1}{5}, P(F)=\frac{2}{5}
  3. CP(E)=25,P(F)=15P(E)=\frac{2}{5}, P(F)=\frac{1}{5}
  4. DP(E)=35,P(F)=45P(E)=\frac{3}{5}, P(F)=\frac{4}{5}

Correct answer

A. P(E)=45,P(F)=35P(E)=\frac{4}{5}, P(F)=\frac{3}{5}

Step-by-step solution

Let E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probability
P(E \cup F)-P(E \cap F)=1125(\frac{11}{25} (i.e. only EE or only FF )<imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252FmBHGp2I9tOtkfP60SSq3iGSoDXC_ovlYuNpy4rtKhKE.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FmBHGp2I9tOtkfP60SSq3iGSoDXC_ovlYuNpy4rtKhKE.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FmBHGp2I9tOtkfP60SSq3iGSoDXC_ovlYuNpy4rtKhKE.original.fullsize.png?alt=media"> \\ Neither of them occurs =225⇒P(Eˉ∩Fˉ)=225=\frac{2}{25}\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25} From Eq. (i), we get P(E)+P(F)-2 P(E \cap F)=1125\frac{11}{25}From Eq. (ii), we get (1−P(E))(1−P(F))=225⇒1−P(E)−P(F)+P(E)\cdotP(F)=225(\begin{gathered} (1-P(E))(1-P(F))=\frac{2}{25}\\ \Rightarrow 1-P(E)-P(F)+P(E) \cdotP(F)=\frac{2}{25} \end{gathered} (iv) From Eqs. (iii) and (iv), we get P(E)+P(F)=75 and P(E)\cdotP(F)=1225∴P(E)⋅{75−P(E)}=1225\begin{aligned}& P(E)+P(F)=\frac{7}{5} \text{ and }P(E) \cdotP(F)=\frac{12}{25}\\ & \therefore \quad P(E) \cdot\left\{\frac{7}{5}-P(E)\right\}=\frac{12}{25} \end{aligned}⇒(P(E))2−75P(E)+1225=0⇒(P(E)−35)(P(E)−45)=0∴P(E)=34 or 45\RightarrowP(F)=45 or 35\begin{aligned}& \Rightarrow \quad(P(E))^2-\frac{7}{5}P(E)+\frac{12}{25}=0 \\ & \Rightarrow \quad\left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0 \\ & \therefore \quad P(E)=\frac{3}{4} \text{ or } \frac{4}{5} \RightarrowP(F)=\frac{4}{5} \text{ or } \frac{3}{5} \end{aligned}$
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