7
$\begingroup$

I was at a job interview and got a question. You have a bag with 100 coins. One of these coins has two heads, other coins are regular. You get one coin from this bag, put it on a table and see head. What the probability that this coin has two heads? Thank you!

$\endgroup$

closed as off-topic by Deusovi, iamwhoiam, Marius, Kevin Cruijssen, JMP Mar 31 '16 at 9:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Deusovi, iamwhoiam, Marius, Kevin Cruijssen, JMP
If this question can be reworded to fit the rules in the help center, please edit the question.

15
$\begingroup$

$2/101$
You see one head among the $101$ possible heads, and $2$ of these heads belongs to the double-head coin so you have $2/101$ chances to have the double-head coin

The probability that you chose the double-head coin is twice the probability that you chose a specific normal coin

Maths explanation :

Let's define :
* $X$ the event "You choose the special coin",
* $H$ the event "you see head"
* $T$ the event "you see tail"

We want to evaluate $P(X|H)$, the probability that we chose the special coin given that we see a head.
$P(X) = P(H)*P(X|H) + P(T)*P(X|T)$
$P(H) = \frac{101}{200}$ ,
$P(T) = \frac{99}{200}$
$P(X) = \frac{1}{100}$
$P(X|T) = 0$ (the coin can't be special if we see tail)
If we substitue :
$\frac{101}{200} P(X|H) = \frac{1}{100}$ so $P(X|H) = \frac{2}{101}$

$\endgroup$
  • $\begingroup$ Where do you get the values for $P(H)$ and $P(T)$? $\endgroup$ – Trenin Mar 31 '16 at 12:11
  • 1
    $\begingroup$ You can also simplify this problem by considering the 100 coins to be 200 coin-faces. So instead of drawing a coin + looking at one side, you perform a single operation: draw one coin-face. Of all coin faces, 101 are heads and 2 of those are the double-heads. $\endgroup$ – florisla Mar 31 '16 at 12:16
  • $\begingroup$ @Trenin There As Florista explained, if you think about all the faces you can choose, there are 101 head and 99 tails and they are all equiprobable. $P(H)=\frac{101}{101+99}$ and $P(T)=\frac{99}{101+99}$ $\endgroup$ – Fabich Mar 31 '16 at 12:20
  • 1
    $\begingroup$ Currently Lord of dark has 666 rep.... I want to up-vote, but don't want you to lose that number. $\endgroup$ – Aggie Kidd Mar 31 '16 at 18:59
  • $\begingroup$ I am the devil ! $\endgroup$ – Fabich Mar 31 '16 at 19:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.