Consider the circuit shown in the image. If t | vedclass

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(NONE) The circuit consists of two parallel branches connected between points $A$ and $B$.$1$. The upper branch contains two switches $a$ and $b$ in s

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$p^2 + p$

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(NONE) The circuit consists of two parallel branches connected between points $A$ and $B$.$1$. The upper branch contains two switches $a$ and $b$ in series. Current flows through this branch only if both switches $a$ and $b$ are closed.Since the probability of each switch being closed is $p$,the probability that the upper branch is closed is $P(	ext{upper}) = p \cdot p = p^2$.$2$. The lower branch contains a single switch $c$. Current flows through this branch if switch $c$ is closed.The probability that the lower branch is closed is $P(	ext{lower}) = p$.$3$. Current flows from $A$ to $B$ if either the upper branch is closed $OR$ the lower branch is closed. Using the inclusion-exclusion principle for probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Here,$P(	ext{upper} \cup 	ext{lower}) = P(	ext{upper}) + P(	ext{lower}) - P(	ext{upper} \cap 	ext{lower})$.$P(	ext{upper} \cap 	ext{lower}) = p^2 \cdot p = p^3$.Therefore,the required probability is $p^2 + p - p^3$.

Consider the circuit shown in the image. If the probability that each switch is closed is $p$,then find the probability of current flowing through $AB$.

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