# Michael has 4 biased coins where each of them comes up head with probability 0.6. Michael plays a game with his 3 friends. Each of them tosses the 4 coins for 5 times. The winner would be the one who gets 3 times at least 3 heads of tosses. ?

## What is the probability that at least three of them would be the winners?

Nov 22, 2017

Assuming the interpretation below is correct, the probability that at least 3 of them "win" is approximately 0.9685.

#### Explanation:

The question is unclear regarding "the winner", however I think the intended meaning is "if someone tosses the 4 coins and gets 3+ heads, they get a point. If they get 3 or more points in 5 tosses, they are called a winner."

Let's call a single toss of all 4 coins a "turn".

Let $X$ be the number of heads in one turn. Then X ~ "BIN"(4, 0.6).

The probability of getting at least 3 heads in one turn is

$\text{P} \left(X \ge 3\right)$
$= \text{P"(X=3) + "P} \left(X = 4\right)$
$= \left(\begin{matrix}4 \\ 3\end{matrix}\right) {\left(0.6\right)}^{3} {\left(0.4\right)}^{1} + \left(\begin{matrix}4 \\ 4\end{matrix}\right) {\left(0.6\right)}^{4} {\left(0.4\right)}^{0}$
$= 0.5184 + 0.1296$
$= 0.648$

So the probability of getting a point on any turn is 0.648.

Let $Y$ be the number of points earned in 5 turns. Then Y~"BIN"(5, 0.648).

The probability of getting 3 or more points in 5 turns is

$\text{P} \left(Y \ge 3\right)$
$= {\sum}_{y = 3}^{5} \text{P} \left(Y = y\right)$
$= {\sum}_{y = 3}^{5} \left(\begin{matrix}5 \\ y\end{matrix}\right) {\left(0.648\right)}^{y} {\left(1 - 0.648\right)}^{5 - y}$
$\approx 0.33714 + 0.31032 + 0.11425$
$\approx 0.7617$

So the probability of one person "winning" is $\approx 0.7617 .$

Let $Z$ be the number of winners out of the 4 friends. Then Z~"BIN"(4,0.7617).

The probability of at least 3 of the friends winning is

$\text{P} \left(Z \ge 3\right)$
$= \text{P"(Z = 3) + "P} \left(Z = 4\right)$
$\approx \left(\begin{matrix}4 \\ 3\end{matrix}\right) {\left(0.7617\right)}^{3} {\left(0.2383\right)}^{1} + \left(\begin{matrix}4 \\ 4\end{matrix}\right) {\left(0.7617\right)}^{4} {\left(0.2383\right)}^{0}$
$\approx 0.6319 + 0.3366$
$= 0.9685$

So the probability of at least 3 out of 4 friends winning is $\approx 0.9685 .$