SATHEE: Probability 3 Question 29

Probability 3 Question 29

29. If two events $A$ and $B$ are such that $P (A^{c}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P [B / (A \cup B^{c})] = \dots$ . $(1994, 2 M)$

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Solution:

$P (A^{c}) = 0.3$

[given]

$\begin{aligned} \Rightarrow P (A) = 0.7 \\ P (B) = 0.4 \\ \Rightarrow P (B^{c}) = 0.6 and P (A \cap B^{c}) = 0.5 \\ Now, P (A \cup B^{c}) = P (A) + P (B^{c}) - P (A \cap B^{c}) \\ ∴ P [B / (A \cup B^{c}] = \frac{P B \cap (A \cup B^{c})}{P (A \cup B^{c})} \\ = 0.7 + 0.6 - 0.5 = 0.8 \\ = \frac{P (B \cap A) \cup (B \cap B^{c})}{0.8} = \frac{P (B \cap A) \cup φ}{0.8} = \frac{P (B \cap A)}{0.8} \end{aligned}$

$\begin{aligned} = \frac{1}{0.8} [P (A) - P (A \cap B^{c})] \\ = \frac{0.7 - 0.5}{0.8} = \frac{0.2}{0.8} = \frac{1}{4} \end{aligned}$

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Probability 3 Question 29

29. If two events A and B are such that P(Ac)=0.3,P(B)=0.4 and P(A∩Bc)=0.5, then P[B/(A∪Bc)]=…. (1994,2M)

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29. If two events $A$ and $B$ are such that $P (A^{c}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P [B / (A \cup B^{c})] = \dots$ . $(1994, 2 M)$