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A fair coin is tossed once. If a head comes up, a fair die is tossed once. If a tail comes up, a fair die is tossed twice. The probability of observing at least one six is:-
I have proceeded like this:-

P(getting one six | heads have come up)= P(getting one six and heads have come up)/P(heads have come up)= (1/6)/(1/2)=1/3

P(one six | tails have come up)=P(one six and tails have come up)/P(tails have come up)= (5/36)/(1/2)= 5/18

P(two six | tails have come up)=P(two six and tails have come up)/P(tails)=(1/36)/(1/2)=1/18

So required probability is 1/18+5/18+1/3 = 2/3
Is this right?

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  • $\begingroup$ If you just roll the die twice, the probability of seeing at least one $6$ is $1-\left( \frac 56\right)^2=\frac {11}{36}$ so your answer is much too high. $\endgroup$ – lulu Feb 8 '17 at 16:57
  • $\begingroup$ How did you decide that P(getting one six and heads have come up) is 1/6? First the coin has to come up heads, and then the die has to be a six. This should be less likely than if you just rolled a die unconditionally and it came up heads. $\endgroup$ – David K Feb 8 '17 at 16:57
  • $\begingroup$ Based on both of your comments I tried this:- P(at least one six)=P(atleast on six | heads)*P(heads) + P(atleast one six | tails)*P(tails) = 1/2*1/6 + 1/2*11/36 =17/72 $\endgroup$ – bandit_king28 Feb 8 '17 at 17:04
  • $\begingroup$ That looks good! $\endgroup$ – lulu Feb 8 '17 at 17:06

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