# There are 3 unfair coins. You pick one of the coins. What is the probability of tossing heads?

There are three coins: c1, c2, and c3. When tossing a coin once, the probabilities of getting a head for c1, c2, and c3 are 0.1, 0.4, and 0.7, respectively. Now suppose that you pick one of the coins, with the probability 0.2 of the coin being c1, 0.5 of being c2, 0.3 of being c3.

1. Now that you have picked a coin, you toss the coin once, what is the probability of getting a head?

2. If you toss this selected coin once and get a head, what is the probability that c1 was the coin selected in the first step? Hint: you can use Bayes’ rule to compute the probability.

For 1), I believe I must combine the probabilities of picking each coin. I don't know if I sum them or if I multiply them. (E.g., 0.2 + 0.5 + 0.3). Then I believe I should combine the probabilities of tossing heads for each coin (again, not sure if I should sum them or multiply them). Then when I have the two numbers associated with both combined probabilities, I think I should multiply them p1 x p2 for my final answer. Can someone help guide me? I don't know how to do 2) either.

1. Use law of total probability getting

$$\mathbb{P}[H]=0.2\times0.1+0.5\times0.4+0.3\times0.7=0.43$$

1. using Bayes' Theorem

$$\mathbb{P}[c_1|H]=\frac{0.2\times0.1}{0.2\times0.1+0.5\times0.4+0.3\times0.7}=\frac{2}{43}$$

• For simplest answer (+1) :-} Commented Oct 5, 2021 at 17:23

The Law of Total Probability says $$\Pr(A) = \Pr(A\cap B)+\Pr(A\cap B^c)$$ and can be generalized further.

Letting $$B_1,B_2,\dots B_n$$ be a partition of the sample space (mutually exclusive and exhaustive) we have that $$\Pr(A)=\sum\limits_{i=1}^n \Pr(A\cap B_i)$$

Using conditional probability, these intersections can be rewritten as $$\Pr(A)=\sum\limits_{i=1}^n\Pr(B_i)\Pr(A\mid B_i)$$

For your problem, let $$B_1$$ be the event the first coin was picked, $$B_2$$ the second, and so on... Let $$A$$ be the event that you flip a head.

We have then...

$$\Pr(A)=\Pr(B_1)\Pr(A\mid B_1)+\Pr(B_2)\Pr(A\mid B_2)+\Pr(B_3)\Pr(A\mid B_3)$$

Each of those numbers above were given in the problem statement.

$$\Pr(A)=0.2\cdot 0.1 + 0.5\cdot 0.4 + 0.3\cdot 0.7$$

As for the second part of the problem, this follows from Bayes' Theorem. $$\Pr(B_1\mid A) = \dfrac{\Pr(A\mid B_1)\Pr(B_1)}{\Pr(A)}$$

Again, each of these numbers were in the problem statement or were calculated in the previous part.

Use the law of total probability, which states that for an event $$A$$ of the sample space and a collection of mutually exclusive and exhaustive events $$E_1, …, E_n$$, the probability of A is given by

$$P(A)=\sum_{k=1}^n P(E_k)P(A|E_k),$$

where each $$P(A|E_k)$$ is the probability that $$A$$ occurs given $$E_k$$.

Let desired event $$A$$ be getting any heads. Let each $$E_k$$ be the probably of picking the $$k$$th coin. These are exhaustive because the sum of their probabilities is $$1$$. Thus, we have

$$P(A)= \sum_{k=1}^3 P(E_k)P(A|E_k).$$

Plugging in the numbers should be straightforward.

Now in $$2)$$ we are asked to find the probability we picked coin $$1$$ given we got a heads. Notationally, this is $$P($$coin 1$$|A)$$. We use the formula for conditional probability. So, we have

$$P(c1|A)=\frac{P(c1 \cap A)}{P(A)}.$$

One calculates the intersection part by using the formula for conditional probability. I will let you do the rest.

In discrete case, it is always good to think in terms of which events are sequential and which are exclusive. Sequential come under multiplication rule and Exclusive come under addition rule.

For this to happen, you need to breakdown your problem into steps. Before getting heads (or flipping the coin), you need to select a coin. You can select either coin 1 or coin 2 or coin 3. Suppose you select coin 1. Then probability of selecting it is $$0.2$$. Now after selecting, you flip it. And then, you will get heads with probability $$0.1$$. Here the event is selecting coin and getting heads. So you see, wherever the events are joined by "and" statement, you need to multiply probability, and wherever the events are joined by an "or" statement, you need to add.

So you problem would be finding probability of

(Selecting coin $$1$$ AND getting heads of coin $$1$$) OR (Selecting coin $$2$$ AND getting heads of coin $$2$$) OR (Selecting coin $$3$$ AND getting heads of coin $$3$$)

= P(Selecting coin $$1$$ AND getting heads of coin $$1$$) + P(Selecting coin $$2$$ AND getting heads of coin $$2$$) + P(Selecting coin $$3$$ AND getting heads of coin $$3$$)

= P(Selecting coin $$1$$)xP(getting heads of coin $$1$$) + P(Selecting coin $$2$$)xP(getting heads of coin $$2$$) + P(Selecting coin $$3$$)xP(getting heads of coin $$3$$)

This is your solution for part $$1$$.

For part $$2$$, simply use Bayes' rule.

Sometimes it helps to think about many trials of the experiment. Say you ran the experiment $$100$$ times. You might reasonably expect the following outcomes:

Coin 1: $$20$$ times with $$2$$ heads and $$18$$ tails.

Coin 2: $$50$$ times with $$20$$ heads and $$30$$ tails.

Coin 3: $$30$$ times with $$21$$ heads and $$9$$ tails.

Thus in $$100$$ trials you would expect to get $$2+20+21=43$$ heads for a probability of $$P({\rm heads})=\frac{43}{100}$$.

For part b), given that you got a heads, you know you are in one of the $$43$$ heads cases, two of which come from flipping coin 1. So Probability of coin 1 given the result of heads is $$\frac{2}{43}$$.