An urn contains $6$ white and $9$ black balls. Two successive draws of $4$ balls are made without replacement. The probability,that the first draw gives all white balls and the second draw gives all black balls,is:

  • A
    $\frac{5}{256}$
  • B
    $\frac{5}{715}$
  • C
    $\frac{3}{715}$
  • D
    $\frac{3}{256}$

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$A$ and $B$ are two events of a random experiment such that $P(B)=0.4$, $P(A \cap \bar{B})=0.5$, and $P(A \cup B) + P\left(\frac{B}{A \cup \bar{B}}\right) = 1.15$. Then $P(A) = $

$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. $A$ coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

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