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Help... having my first encounter on conditional probability and then I meet this problem.. it includes several conditions and I am not sure if Im thinking the right procedure on this problem. Any idea will be of great help. Thanks in advance.

There are 3 coins. One is two headed coin, the other one comes heads 75 percent of the time and the third is a fair coin. A coin is selected at random and flipped.

1) What is the probability that the flipped coin will come up head?

2) Given that the coin that was flipped comes up head, what is the probability that it was fair coin?

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  • $\begingroup$ make some tree diagram... separating each of three conditions...then another branches for successes and fails ,,, then I got confused if im still doing the right thing... $\endgroup$ – rosa Mar 3 '17 at 10:16
  • $\begingroup$ Think of it as what contribution does the fair coin have in the head sample space? $\endgroup$ – samjoe Mar 3 '17 at 10:27
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For the first, simply use the basic definition of probability

$$\dfrac{Favourable}{Total}=\frac{1+\frac{3}{4}+\frac{1}{2}}{3}$$

For the second, use Bayes Theorem. Let the events be :

  1. $U_1$: Coin with both heads flipped.

  2. $U_2$: Coin which shows head 3/4 times flipped.

  3. $F$: Fair coin flipped.

  4. $H$ : Head is the result.

Then we need to find $P(F|H)$

$$P(F|H) = \dfrac{P(H|F)\cdot P(F)}{P(H|U_1)\cdot P(U_1)+P(H|U_2)\cdot P(U_2)+P(H|F)\cdot P(F)}$$

$$=\dfrac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{1}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}} = 2/9$$

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  • $\begingroup$ aw! the second solution seems tedious hahaha!!!.. yet I'll try my best to understand your answer tnx a lot sir! $\endgroup$ – rosa Mar 3 '17 at 10:28
  • $\begingroup$ @rosa No its not at all tedious! Rather you'll find it more easy than making tree in complex problems.. $\endgroup$ – samjoe Mar 3 '17 at 10:29
  • $\begingroup$ Im trying to understand the reason how u come up to such formula haha :D a lot of tnx sir, really :) $\endgroup$ – rosa Mar 3 '17 at 10:31
  • $\begingroup$ @rosa $\def\P{\mathop{\mathsf P}}\P(F\mid H)= \dfrac{\P(H\mid F)\P(F)}{\P(H)}$ is by Bayes' Rule. $\P(H)=\P(H\mid U_1)\P(U_1)+\P(H\mid U_2)\P(U_2)+\P(H\mid F)\P(F)$ is from the Law of Total Probability. They are often combined. $\endgroup$ – Graham Kemp Mar 3 '17 at 10:52
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For a)

Just create a tree diagram with all the different options: then, follow the tree to find the probability of each event. It should equal 3/4

B is a little bit more complex

The formula for conditional probability is P(A|B) = P(A and B)/P(B)

In this case

The probability of a fair coin, if it is known that the outcome is a heads.

The probability of getting a heads is 3/4 as we solved for in a:

so we get P(A and B)/0.75

The probability that you choose the fair coin is 1/3, and the probability of heads on said coin is 1/2

So, probability of A and B is 1/6

if you do (1/6)/(3/4)

you get 2/9

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Using tree diagram, a) P[H] = P[C1H] + P[C2H]+P[C3H]= 1/3 + 1/4 + 1/6 = 3/4 b) P[C3|H] = P[C3H]/P[H] = (1/6)/ (3/4) = 2/9

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  • $\begingroup$ Your calculations look correct, but may I recommend a couple of improvements. First, your answer would be better if you explained your notation $C_1H$, etc. Second, please use MathJax to format your answer. $\endgroup$ – Jeremy Dover Feb 21 at 13:29

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