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Consider a three level tree diagram of events happening in a sequence.

At first level $A, B$ or $C$ events are possible, at second level $D, E$ or $F$ are possible and at third level $P$ or $Q$ are possible.

All edge probabalties at first level are $1 / 3,$ at second level are $1/3,$ and at third level are $1/2.$

Find $P(A \mid P)?$

  1. $2/3$
  2. $1 / 3$
  3. $1 / 2$
  4. $3 / 4$
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12 votes

We have to find $P(A|P)$

$\therefore P(A | P) = \frac{P(A \cap P)}{P(P)}$

So, $P(P)$ =  $9 \times [(\frac{1}{3}) \times (\frac{1}{3}) \times (\frac{1}{2})] = \frac{9}{18}$

$P(A \cap P) = P(A \cap D \cap P) + P(A \cap E \cap P) + P(A \cap F \cap P) = 3 \times [(\frac{1}{3}) \times (\frac{1}{3}) \times (\frac{1}{2})] = \frac{3}{18}$

So, $P(A|P) = \frac{\frac{3}{18}}{\frac{9}{18}} = \frac{1}{3}$

$Ans: B$

2 votes
2 votes
$P(A/P) = P(A\cap P)\div P(P) = [ 3\ast (1/3)\ast (1/3)\ast (1/2)]\div 9/18 =1/3$
Answer:

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