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Question . A coin box contains 8 fair, standard coins (head and tails) and 1 coin which has heads on both sides. A boy selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?

My approach:

Let “Hi” be the event that Head occurs at the ith flip.. i=1(1)5

So the question asks for the conditional probability,

P(H5|H1, H2, H3, H4)

=P(H1, H2, H3, H4, H5)/P(H1, H2, H3, H4)

=P(H1)P(H2)P(H3)P(H4)P(H5)/P(H1)P(H2) P(H3)P(H4)

[The events “Hi’s” are independent]

=P(H5)

=P(H)

[Where, “H” is the event of occurance of head]

=[P(H|fair coin selected)×P(fair coin selected)]+[P(H|unfair coin selected)×P(unfair coin selected)] [By theorem of Total Probability]

=[(1/2)×(8/9)]+[(1)×(1/9)]

=5/9

My confusion:

My problem is that for a coin that gives 4 heads as outcome, it is very probable that it is the biased coin and if I use this Classical definition, it is somehow not reflecting in the resultant probability of getting the 5th head..It is 5/9 which is near to 0.556.

But as getting 4 heads previously indicates high possibility of selectiin of biased coin, the 5th toss getting head should have a probability figure close to 1.

Can you state any other convenient way to solve this problem ?

Thank you.

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    $\begingroup$ Consider an extreme case where there are 2 coins in the box: one with two heads and one with two tails. Would you still write that $P(H1,\dots,H5)=P(H1)\cdot \dots \cdot P(H5) $? Probably not, the former is 1/2 while the later is 1/32. $\endgroup$ – Luca Citi Apr 7 at 6:10
  • $\begingroup$ Yes I get you...But here actually I wanted to select a coin 1st..Then flip it..Say I've selected an unbiased coin..Now I'm flipping it 5 times..Then the flips are independent..Here "Hi" denotes that I am getting heads from that particular coin that I've selected..And probability of getting any particular coin(among 8 unbiased and 1 biased) is shown by the "total probability" at the end..Am I wrong? $\endgroup$ – P db Apr 7 at 6:28
  • $\begingroup$ Actually my answer should have been upside down..I should have selected the coin 1st by total probability..Then put the conditional probability.. $\endgroup$ – P db Apr 7 at 6:33
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Heads are independent given the coin is fair or biased, i.e. $P(H_2,H_1|F)=P(H_2|F)P(H_1|F)$. So, throws are conditionally independent. Therefore, we have:

$$P(H_{1-5})=P(H_{1-5}|F)P(F)+P(H_{1-5}|F')P(F')=\frac{1}{32}\frac{8}{9}+\frac{1}{9}=\frac{5}{36}$$

Similarly, $$P(H_{1-4})=P(H_{1-4}|F)P(F)+P(H_{1-4}|F')P(F')=\frac{1}{16}\frac{8}{9}+\frac{1}{9}=\frac{1}{6}$$

So, $P(H_5|H_{1-4})=5/6\approx 0.83$

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