JEE Advanced 2012Paper 1
- Exam
- JEE Advanced
- Year
- 2012
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Probability
JEE Advanced 2012 · Paper 1 · Mathematics · Probability
A ship is fitted with three engines E_ 1 , E_ 2 and E_ 3 . The...
A ship is fitted with three engines and . The engines function independently of each other with respective probabilities and . For the ship to be operational at least two of its engines must function. Let denote the event that the ship is operational and let and denote respectively the events that the engines and are functioning. Which of the following is(are) true?
Options
- AP
- B [Exactly two engines of the ship are functioning
- CP
- DP
Correct answer
B. [Exactly two engines of the ship are functioning
Step-by-step solution
Given that (at least 2 engines are functioning)
=\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{4}P\left(X_{1}^{C} / X\right)=\frac{P\left(X_{1}^{C} \cap X\right)}{P(X)}=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)}{P(X)} =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8} \thereforeP/ X] =\frac{P[(\text{ Exactly two engines are functioning }) \cap X]}{P(X)} =\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P(X)} =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8} \thereforeP\left(X / X_{2}\right)=\frac{P\left(X \cap X_{2}\right)}{P\left(X_{2}\right)} =\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{2}\right)} =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8} \thereforeP\left(X / X_{1}\right)=\frac{P\left(X \cap X_{1}\right)}{P\left(X_{1}\right)} =\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{1}\right)} =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16} \therefore$ (d) is true.