An electronic assembly consists of two subsystems,$A$ and $B$. From previous testing procedures,the following probabilities are known:
$P(A \text{ fails}) = 0.2$
$P(B \text{ fails alone}) = 0.15$
$P(A \text{ and } B \text{ fail}) = 0.15$
Evaluate the probability $P(A \text{ fails } | \text{ } B \text{ has failed})$.
$P(A \text{ fails}) = 0.2$
$P(B \text{ fails alone}) = 0.15$
$P(A \text{ and } B \text{ fail}) = 0.15$
Evaluate the probability $P(A \text{ fails } | \text{ } B \text{ has failed})$.
- A$0.5$
- B$0.4$
- C$0.3$
- D$0.2$
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Similar Questions
If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{3}$,then the value of $P(A^{\prime} | B^{\prime}) + P(B^{\prime} | A^{\prime})$ is:
If $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$,find $P(A | B)$.
Easy
View SolutionIf two unbiased dice are rolled simultaneously until a sum of the numbers appearing on these dice is either $7$ or $11$, then the probability that $7$ comes before $11$ is:
Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:
List-$I$ List-$II$ $(A)$ $P(E_2)$ $(i)$ $1/4$ $(B)$ $P(E_1 \cup E_2)$ $(ii)$ $5/8$ $(C)$ $P(\bar{E}_1 / \bar{E}_2)$ $(iii)$ $1/8$ $(D)$ $P(E_1 / \bar{E}_2)$ $(iv)$ $1/2$ $(v)$ $3/8$ $(vi)$ $3/4$
| List-$I$ | List-$II$ |
| $(A)$ $P(E_2)$ | $(i)$ $1/4$ |
| $(B)$ $P(E_1 \cup E_2)$ | $(ii)$ $5/8$ |
| $(C)$ $P(\bar{E}_1 / \bar{E}_2)$ | $(iii)$ $1/8$ |
| $(D)$ $P(E_1 / \bar{E}_2)$ | $(iv)$ $1/2$ |
| $(v)$ $3/8$ | |
| $(vi)$ $3/4$ |
DifficultAP EAMCET 2009
View SolutionIf $A$ and $B$ are events such that $P(A) > 0$ and $P(B) \neq 1$,then $P(A \mid B^{\prime}) = $ . . . . . . .
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