$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have heads on both sides,whereas the other $n+1$ coins are fair. One coin is selected at random and tossed. If the probability that the toss results in heads is $\frac{31}{42}$,then the value of $n$ is

  • A
    $6$
  • B
    $8$
  • C
    $10$
  • D
    $5$

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Similar Questions

Coloured balls are distributed in four boxes as shown in the following table:
Box Black White Red Blue
$I$$3$$4$$5$$6$
$II$$2$$2$$2$$2$
$III$$1$$2$$3$$1$
$IV$$4$$3$$1$$5$

$A$ box is selected at random and then a ball is randomly drawn from the selected box. If the colour of the ball is black,what is the probability that the ball drawn is from box $III$?

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Three boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:
Box White,Black,Red
$B_1$ $2, 1, 2$
$B_2$ $3, 2, 4$
$B_3$ $4, 3, 2$

$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is

$A$ person is known to speak truth in $3$ out of $5$ times. If he throws a die and reports that it is six,then the probability that it is actually six,is

On every evening,a student either watches $TV$ or reads a book. The probability of watching $TV$ is $\frac{4}{5}$. If he watches $TV$,the probability that he will fall asleep is $\frac{3}{4}$ and it is $\frac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening,the probability that he watched the $TV$ is

Four boxes $A, B, C$ and $D$ contain $5000, 3000, 2000$ and $1000$ fuses respectively. The percentages of defective fuses in these boxes are $3\%, 2\%, 1\%$ and $0.5\%$ respectively. If a fuse selected at random from one of the boxes is found to be defective, then the probability that it has come from box $D$ is

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