0 like 0 dislike
1,718 views
I have in my pocket ten coins. Nine of them are ordinary coins with equal chances of coming up head and tail when tossed and the tenth has two heads.

(a) If I take one of the coins at random from my pocket, what is the probability that it is the coin with two heads?

(b) If I toss the coin and it comes up heads, what is the probability that it is the coin with two heads ?

(c) If I toss the coin one further time and it comes up tails, what is the probability that it is one of the nine ordinary coins ?
| 1,718 views

0 like 0 dislike
Denote by $D$ the event that the coin is the one with two heads.
(a) $\mathrm{P}(D)=1 / 10$.

(b) By Bayes theorem, we have
$$\mathrm{P}(D \mid H)=\frac{\mathrm{P}(H \mid D) \mathrm{P}(D)}{\mathrm{P}(H)}$$
We have $\mathrm{P}(H \mid D)=1$ and $\mathrm{P}(D)=\frac{1}{10}$. Now, we need to think about $H$, getting a head, in terms of getting a head with either a double headed or single headed coin. Using the idea of a partition,
\begin{aligned} \mathrm{P}(H) &=\mathrm{P}(H \cap D)+\mathrm{P}\left(H \cap D^{c}\right) \\ &=\mathrm{P}(H \mid D) \mathrm{P}(D)+\mathrm{P}\left(H \mid D^{c}\right) \mathrm{P}\left(D^{c}\right) \\ &=(1)\left(\frac{1}{10}\right)+\left(\frac{1}{2}\right)\left(\frac{9}{10}\right) \\ &=\frac{11}{20} \end{aligned}
Finally, here is another way of calculating $\mathrm{P}(H):$ think of the bag as containing the possible tosses. As the bag contains 9 fair coins and one double-headed coin, it must contains 11 heads and 9 tails, so that the probability of choosing a head is $11 /(11+9)=11 / 20$. To return to the original question, we now obtain the answer
$$\mathrm{P}(D \mid H)=\frac{\frac{1}{10}}{\frac{11}{20}}=\frac{2}{11}$$
(c) 1. If it comes up tails, it can't be the coin with two heads. Therefore it must be one of the other nine.
by Platinum (142,760 points)

0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike