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Question is

Alice tosses a fair coin, then rolls a fair die. If the result is <=2, she tosses fair coin ,otherwise she tosses an unfair coin(80% gets head). .what is the probability that she gets exactly one H in two tosses?

Here's my thought. I want one h in two tosses, so first tosses she got 1/2 H or 1/2 T. If she got H, second tosses has to be T. And in second tosses, she gets 2/6 to have fair coin, and 4/6 chances to get unfair coin.

So toss H first, 1/2* (2/6*1/2+4/6*0.2)=0.15

If toss T first,we want second time to be H, 1/2*(2/6*1/2+4/6*0.8)=0.35

So the probability for question is 0.5?

Am I right?

It feels like i was wrong...

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$\checkmark$ That looks right.

Let $H_1,H_2$ be the event of rolling a head on the first or second toss, and let $F$ be the event of tossing a fair coin on the second toss as determined by the roll of a fair die. The first toss is always fair.

$$\begin{align} & \quad\; \mathsf P((H_1\cap H_2^c)\cup(H_1^c\cap H_2)) \\[2ex] & = \mathsf P(H_1)\Big(\mathsf P(H_2^c\mid F)\mathsf P(F)+\mathsf P(H_2^c\mid F^c)\mathsf P(F^c)\Big) + \mathsf P(H_1^c)\Big(\mathsf P(H_2\mid F)\mathsf P(F)+\mathsf P(H_2\mid F^c)\mathsf P(F^c)\Big) \\[2ex] & = \frac 1 2\Big(\frac 1 2\frac 1 3+\frac 1 5\frac 2 3\Big) +\frac 1 2\Big(\frac 1 2\frac 1 3+\frac 4 5\frac 2 3\Big) \\[2ex] & = \frac 1 2 \end{align}$$

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  • $\begingroup$ Thank you so much! That equation is really helpful. I actually have problem to list the equation every time, coz my logic is not so clear. Thank you! $\endgroup$ – Amanda_Q Nov 21 '14 at 2:48

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