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An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
$\mathrm{P}(\mathrm{A}$ fails $)=\mathbf{0 . 2}$
$P(B$ fails alone $)=0.15$
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B}$ fail $)=\mathbf{0 . 1 5}$
Evaluate the following probabilities
(i) $\mathbf{P}(\mathbf{A}$ fails $\mid \mathrm{B}$ has failed)
(ii) $\mathbf{P}$ (A fails alone)
$\mathrm{P}(\mathrm{A}$ fails $)=\mathbf{0 . 2}$
$P(B$ fails alone $)=0.15$
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B}$ fail $)=\mathbf{0 . 1 5}$
Evaluate the following probabilities
(i) $\mathbf{P}(\mathbf{A}$ fails $\mid \mathrm{B}$ has failed)
(ii) $\mathbf{P}$ (A fails alone)
Solution:
1105 Upvotes
Verified Answer
Events A fails, and B fails are denoted by A and B respecitvely
We have $P(A)=0.2$ and $P(A$ and $B$ fail $)=0.15$
i.e., $P=(\bar{A} \cap \bar{B}) 0 \cdot 15$
$\mathrm{P}(\overline{\mathrm{B}}$ alone $)=\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
Now $0.15=\mathrm{P}(\overline{\mathrm{B}})-0.15 \quad \therefore \mathrm{P}(\overline{\mathrm{B}})=0.30$
(i) $\quad \mathrm{P}=(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\frac{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}{(\overline{\mathrm{B}})}$
$$
=\frac{0 \cdot 15}{0 \cdot 30}=\frac{1}{2}=0.5
$$
(ii) $\mathrm{P}(\mathrm{A}$ fails alone $)=\mathrm{P}(\mathrm{A}$ alone $)$
$$
=\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=0.20-0.15=0-05
$$
We have $P(A)=0.2$ and $P(A$ and $B$ fail $)=0.15$
i.e., $P=(\bar{A} \cap \bar{B}) 0 \cdot 15$
$\mathrm{P}(\overline{\mathrm{B}}$ alone $)=\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
Now $0.15=\mathrm{P}(\overline{\mathrm{B}})-0.15 \quad \therefore \mathrm{P}(\overline{\mathrm{B}})=0.30$
(i) $\quad \mathrm{P}=(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\frac{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}{(\overline{\mathrm{B}})}$
$$
=\frac{0 \cdot 15}{0 \cdot 30}=\frac{1}{2}=0.5
$$
(ii) $\mathrm{P}(\mathrm{A}$ fails alone $)=\mathrm{P}(\mathrm{A}$ alone $)$
$$
=\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=0.20-0.15=0-05
$$
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