# Probability of Two Coins (one fair and one biased)

Two coins are in a hat. The coins look alike, but one coin is fair (with probability 1/2 of Heads), while the other coin is biased, with probability 1/4 of Heads. One of the coins is randomly pulled from the hat, without knowing which of the two it is. Call the chosen coin “Coin C”

• Find the probability that in 10 flips of Coin C, there will be exactly 3 Heads. (The coin is equally likely to be either of the 2 coins; do not assume it already landed Heads twice as in (a).)

I have the following answers as shown below but I am unsure what the 10 3 in brackets is? Thanks in advance

Let $$X$$ be the number of Heads in 10 tossess. By the Law of Total Probability (conditioning on which of the two coins is C)\begin{aligned}\mathsf P(X=3)~&=~\mathsf P(X=3\mid\text{fair})\,\mathsf P(\text{fair})+\mathsf P(X=3\mid\text{biased})\,\mathsf P(\text{biased})\\[2ex]&=~\dbinom{10}3\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac 12+ \dbinom{10}3\left(\dfrac{1}{4}\right)^3\left(\dfrac{3}{4}\right)^7\cdot\dfrac{1}{2}\\[2ex]&=~\dfrac 12\dbinom{10}3\left(\dfrac 1{2^{10}}+\dfrac{3^7}{4^{10}}\right)\end{aligned}

• The "answers as shown below" aren't attached. Could you please attach them? Aug 22 '21 at 5:31
• Calculate probability for each coin and then take the average Aug 22 '21 at 5:36
• i.stack.imgur.com/2XX25.png is the answers Aug 22 '21 at 5:57

The binomial coefficient, $$\tbinom{10}3$$ is the count of distinct ways to select $$3$$ items from a set of $$10$$. Sometimes written as $$^{10}\mathrm C_3$$, it equals $$\tfrac{10!}{3!~7!}$$.
As such, it is the count for ways $$3$$ heads (and $$7$$ tails) may occur among the $$10$$ flips of coin C. (IE: ways to select 3 from 10 flips to become the heads.)