1
$\begingroup$

I just started studying probability. I am trying to solve this exercise:

When coin A is flipped it comes up heads with probability 1/4, whereas when coin B is flipped it comes up heads with probability 3/4. Suppose that one of these coins is randomly chosen and is flipped twice. If both flips land heads, what is the probability that coin B was the one flipped?

First I have that $A$ is the event in which the coin A is chosen. $B = A'$ is the event in which coin B is chosen.

So I have that $P(A) = P(B) = 1/2$

I also have that $H$ is the event in which the coin lands heads.

I compute $P(H)$ using Bayes' Rule.

$P(H) = P(H|A)P(A) + P(H|B)P(B) = (\frac14) (\frac12) + (\frac34) (\frac12) = \frac12$

As these are independent events, I can say that $P(HH) = P(H)P(H) = \frac14$ is the probability of having 2 tails.

Then I need to compute $P(B|HH)$ to solve the exercise.

I have this:

$P(B|HH) = \frac{P(BHH)}{P(HH)} = \frac{P(B)P(H|B)P(H|BH)}{P(HH)} $

I got that last part using the multiplication rule. Intuitively I would say that $P(H|BH) = P(H|B) $, so using that I can substitute everything and compute the actual probability.

$P(B|HH) = \frac{P(B)P(H|B)P(H|BH)}{P(HH)} = \frac{ \frac12 \frac34 \frac34 }{\frac14} = \frac{\frac9{32}}{\frac14} = \frac{9}{8} \gt 1$

Obviously I am doing something wrong, but I can't see what :(

$\endgroup$
4
$\begingroup$

The probability of getting two heads is

$$\frac34\cdot\frac34=\frac9{16}$$

if you chose coin $B$ and

$$\frac14\cdot\frac14=\frac1{16}$$ if you chose coin $A$. Thus,

$$\begin{align*} P(HH)&=\frac12\cdot\frac9{16}+\frac12\cdot\frac1{16}=\frac5{16}\;,\\\\ P(B\text{ and }HH)&=\frac12\cdot\frac9{16}=\frac9{32}\;,\text{ and}\\\\ P(B\mid HH)&=\frac{P(B\text{ and }HH)}{P(HH)}=\frac{9/32}{5/16}=\frac9{10}\;. \end{align*}$$

$\endgroup$
  • $\begingroup$ Thanks! I should pay more attention with independent events. $\endgroup$ – Sofia Ontiveros Jun 2 '15 at 1:19
  • $\begingroup$ @Sofia: You’re welcome! $\endgroup$ – Brian M. Scott Jun 2 '15 at 1:20
1
$\begingroup$

Because the same coin is flipped twice, the events are only conditionally independent.

This is true: $P(H_1, H_2\mid A) = P(H_1\mid A)P(H_2\mid A) = (\frac 1 4)^2$

As is this: $P(H_1, H_2\mid B) = P(H_1\mid B)P(H_2\mid B) = (\frac 3 4)^2$

However: $P(H_1, H_2) \neq P(H_1)P(H_2)$

So we use the law of total probability:

$$\begin{align} \mathsf P(H_1,H_2) & = \mathsf P(A)\mathsf P(H_1,H_2\mid A)+\mathsf P(B)\mathsf P(H_1,H_2\mid B) \\[4ex] \mathsf P(B\mid H_1,H_2) & = \frac{\mathsf P(B)\mathsf P(H_1,H_2\mid B)}{\mathsf P(A)\mathsf P(H_1,H_2\mid A)+\mathsf P(B)\mathsf P(H_1,H_2\mid B)} \\ & = \frac{9/16}{1/16+9/16} \\ &= \frac 9 {10} \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.