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STD 12 - 13. probability — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 13. probability1 Mark
MCQ
A ship is fitted with three engines $E_1, E_2$ and $E_3$. The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{4}$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X _1, X _2$ and $X _3$ denotes respectively the events that the engines $E_1 E_2$ and $E_3$ are functioning. Which of the following is (are) true?

$(A)$ $P\left[X_1^c \mid x\right]=\frac{3}{16}$

$(B)$ $P [$ Exactly two engines of the ship are functioning $\mid X ]=\frac{7}{8}$

$(C)$ $P\left[X \mid X_2\right]=\frac{5}{16}$

$(D)$ $P\left[X \mid X_1\right]=\frac{7}{16}$

  • ✓
    $(B,D)$
  • B
    $(B,C)$
  • C
    $(A,D)$
  • D
    $(C,D)$
✓

Answer

Correct option: A.
$(B,D)$
a
$P\left(x_1\right)=\frac{1}{2} $

$P\left(x_2\right)=\frac{1}{4} $

$P\left(x_3\right)=\frac{1}{4} $

$P(x)=P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1 \bar{E}_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right) $

$=\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} $

$P(x)=\frac{1}{4}$

$(A)$ $P\left(\frac{x_1^c}{x}\right)=\frac{P\left(x_1^c \cap x\right)}{P(x)}$

$=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$

$(B)$ $P$ ( exactly two / $x$)$=\frac{P(\text { exactly two } \cap x)}{P(x)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8}$

$(C)$ $P\left(x / x_2\right)=\frac{P\left(x \cap x_2\right)}{P\left(x_2\right)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8}$

$(D)$ $P\left(x / x_1\right)=\frac{P\left(x \cap x_1\right)}{P\left(x_1\right)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$

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