A ship is fitted with three engines E_1, E_2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given probabilities: $P(X_1) = \frac{1}{2}, P(X_2) = \frac{1}{4}, P(X_3) = \frac{1}{4}$.Let $X_1^c, X_2^c, X_3^c$ be the events that engines do no

Vedclass · Accepted Answer

$(B, D)$

Vedclass · Answer

(A) Given probabilities: $P(X_1) = \frac{1}{2}, P(X_2) = \frac{1}{4}, P(X_3) = \frac{1}{4}$.Let $X_1^c, X_2^c, X_3^c$ be the events that engines do not function,so $P(X_1^c) = \frac{1}{2}, P(X_2^c) = \frac{3}{4}, P(X_3^c) = \frac{3}{4}$.The ship is operational $(X)$ if at least two engines function:$P(X) = P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3) + P(X_1 X_2 X_3)$$P(X) = (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}) + (\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}) + (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}) + (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}) = \frac{3}{32} + \frac{3}{32} + \frac{1}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$.$(A) P(X_1^c \mid X) = \frac{P(X_1^c \cap X)}{P(X)} = \frac{P(X_1^c X_2 X_3)}{P(X)} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{1}{8} 
eq \frac{3}{16}$. (False)$(B) P(	ext{Exactly two} \mid X) = \frac{P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3)}{P(X)} = \frac{\frac{3}{32} + \frac{3}{32} + \frac{1}{32}}{\frac{1}{4}} = \frac{7/32}{1/4} = \frac{7}{8}$. (True)$(C) P(X \mid X_2) = \frac{P(X \cap X_2)}{P(X_2)} = \frac{P(X_1 X_2 X_3^c) + P(X_1^c X_2 X_3) + P(X_1 X_2 X_3)}{P(X_2)} = \frac{\frac{3}{32} + \frac{1}{32} + \frac{1}{32}}{\frac{1}{4}} = \frac{5/32}{1/4} = \frac{5}{8} 
eq \frac{5}{16}$. (False)$(D) P(X \mid X_1) = \frac{P(X \cap X_1)}{P(X_1)} = \frac{P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1 X_2 X_3)}{P(X_1)} = \frac{\frac{3}{32} + \frac{3}{32} + \frac{1}{32}}{\frac{1}{2}} = \frac{7/32}{1/2} = \frac{7}{16}$. (True)

Similar Questions

If $A$ and $B$ are two independent events such that $P(\bar{A})=0.75$,$P(A \cup B)=0.65$ and $P(B)=x$,then find the value of $x$.

$A$ dice marked with digits $\{1, 2, 2, 3, 3, 3\}$ is thrown three times. The probability that the sum of the numbers on the faces is $6$ is equal to:

$A$ die is rolled three times. The probability of getting their sum equal to a prime number of the form $4n+1$ is

Consider a system of equations $ax + by = 0$ and $cx + dy = 0$,where $a, b, c, d \in \{0, 1\}$.
Statement $-1$: The probability that the system of equations has a solution is $1$.
Statement $-2$: The probability that the system of equations has a unique solution is $\frac{3}{8}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module