$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $
$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $
- A
$P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C)$
- B
$P(A) + P(B) + P(C) - P(B)\,P(C)$
- C
$P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$
- D
None of these
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If $P\,(A) = \frac{1}{4},\,\,P\,(B) = \frac{5}{8}$ and $P\,(A \cup B) = \frac{3}{4},$ then $P\,(A \cap B) = $
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Two aeroplanes $I$ and $II$ bomb a target in succession. The probabilities of $l$ and $II$ scoring a hit correctlyare $0.3$ and $0.2,$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is
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- [AIEEE 2007]
Fill in the blanks in following table :
$P(A)$
$P(B)$
$P(A \cap B)$
$P (A \cup B)$
$0.35$
...........
$0.25$
$0.6$
Fill in the blanks in following table :
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
| $0.35$ | ........... | $0.25$ | $0.6$ |
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$