The odds in favour of getting a sum that is a multiple of $3$,when a pair of dice is thrown,is:

  • A
    $1: 2$
  • B
    $2: 3$
  • C
    $1: 1$
  • D
    $3: 4$

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If $A$ and $B$ are two events such that $P(A \cup B) + P(A \cap B) = \frac{7}{8}$ and $P(A) = 2P(B)$, then $P(A) = $

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$A, B, C$ are three events,one of which must and only one can happen. The odds in favour of $A$ are $4:6$,and the odds against $B$ are $7:3$. Then,the odds against $C$ are:

For an event,the odds against are $6 : 5$. The probability that the event does not occur is

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