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When flipped, the three coins come up heads with probability 0.5, 0.3, 0.6 respectively. Suppose that I pick one of these three coins entirely at random and flip it three times. 1. What is P(HTT)? (i.e., it comes up heads on the first flip and tails on the last two flips.) 2.Assuming that the three flips, in order, are HTT, what is the probability that the coin that I picked was the fair coin? Don't need to reduce fractions

Work: 1. ((.5*.5)/(.5*.5))/3 + ((.3*.5)/(.7*.5))/3+ ((.6*.5)/(.4*.5))/3 - I think this is wrong 2. I dont know how to do

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    $\begingroup$ Hint: use Bayes rule $\endgroup$ – Jonathan D Feb 15 '18 at 23:39
  • $\begingroup$ I know its conditional to heads tails tails with the different probability for a one in three chance of picking a coin of different probabilities. I am confused on how to set up Bayes rule for that. $\endgroup$ – White Mamba Feb 15 '18 at 23:44
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    $\begingroup$ You have to divide the probability of obtaining HTT with the fair coin by the total probability of obtaining HTT with one of the coins. Please edit your question to show what you have attempted and explain where you are stuck. $\endgroup$ – N. F. Taussig Feb 15 '18 at 23:47
  • $\begingroup$ I am confused on how to go about this problem $\endgroup$ – White Mamba Feb 15 '18 at 23:55
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P(HHT) for each coin.

Coin 1 $P_1 = 0.5^3$

Coin 2 $P_2 = 0.3\cdot 0.7^2$

Coin 3 $P_2 = 0.6\cdot 0.4^2$

$P(HTT) = \frac 13 P_1 + \frac 13P_2 +\frac 13P_3$

$P(coin1|HHT) = \frac {P_1}{P_1+P_2+P_3}$

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  • $\begingroup$ Do I need to do P1/3 on the last part? also do 1/3 for P(HTT) on the last part $\endgroup$ – White Mamba Feb 16 '18 at 0:24
  • $\begingroup$ thank you for your help $\endgroup$ – White Mamba Feb 16 '18 at 0:42
  • $\begingroup$ i.e.: $P(coin1|HHT)=P(coin1 \, and\, HHT)/P(HHT)=1/3*P_1/P(HHT)=P_1/(P_1+P_2+P_3)$ $\endgroup$ – G Cab Feb 16 '18 at 0:48
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To set up Bayes' rule, first think the following way : FIX a coin. Now look at the probability of success.

Suppose a coin has probability $p$ of coming heads. From independence of successive flips, we conclude that the probability of the first three flips looking like $HTT$ is $p(1-p)^2$.

Now, you have three such coins, with $p=0.6,0.5,0.3$. So we can calculate the probability of , each one being FIXED, the first three flips looking like HTT, and these are (easy to see) respectively $0.096,0.125,0.147$.

Now, one of these coins are chosen uniformly at random. So, let us call the coins as coin $1,2,3$ in the order $p=0.6,0.5,0.3$.

Now, Bayes' rule would basically say this : the probability that you get HTT is the sum of the following:

1) The probability that you choose coin $1$, and it comes HTT.

2) The probability that you choose coin $2$, and it comes HTT.

3) The probability that you choose coin $3$, and it comes HTT.

What are these? Well, the choice is independent of the flipping, and is uniform, hence we have the probability of each one being chosen is $\frac 13$. Finally, the complete answer is $\frac{0.096+0.125+0.147}{3} = \frac{46}{375}$.

Now, the second question is asking this : given that HTT has occurred, what is the probability that we chose coin $2$. But then, coin $2$ is just case $2$ above, which is just the probability that with the second coin fixed, we get HTT.

So the answer is just the probability of getting HTT with coin $2$, divided by the probability of getting HTT with any of them(but here, there is no choice to be made, so we do not need to multiply by $\frac 13$). The numerator is then just $0.125 = \frac 18$ while the denominator is $\frac {46 \times 3}{375}$. Now, you can simplify and get $\frac{375}{1104}$.

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  • $\begingroup$ dont you have to multiple each one by 1/3 for part 1? which gives you .156? $\endgroup$ – White Mamba Feb 16 '18 at 0:31
  • $\begingroup$ I did that, if you notice. I did do : $\frac13 \times 0.096 + \frac 13 \times 0.125 + \frac 13 \times 0.147$ above. Only thing was, since $\frac 13$ is common, I used to distributive law to write this as $\frac{0.096+0.125+0.147}{3}$, and that gives the answer $\frac{46}{375}$. You can check that this, and not $0.156$ below, is correct. $\endgroup$ – Teresa Lisbon Feb 16 '18 at 0:38
  • $\begingroup$ Thank you for your help $\endgroup$ – White Mamba Feb 16 '18 at 0:42
  • $\begingroup$ You are welcome! But your username? $\endgroup$ – Teresa Lisbon Feb 16 '18 at 0:45
  • $\begingroup$ NBA player nickname google.com/… $\endgroup$ – White Mamba Feb 16 '18 at 16:33
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First: $$\frac13\cdot \frac18+\frac13\cdot 0.3\cdot 0.7^2+\frac13\cdot 0.6\cdot 0.4^2=0.122667.$$ Second: $$\frac{\frac13\cdot \frac18}{0.122667}=\frac{0.041667}{0.122667}=0.3397.$$

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