# 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. ?

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

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

##### 1 Answer

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

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

#"P"(X >= 3)#

#="P"(X=3) + "P"(X=4)#

#=((4),(3))(0.6)^3(0.4)^1+((4),(4))(0.6)^4(0.4)^0#

#=0.5184+0.1296#

#=0.648#

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

Let

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

#"P"(Y >= 3)#

#=sum_(y=3)^5"P"(Y=y)#

#=sum_(y=3)^5((5),(y))(0.648)^y(1-0.648)^(5-y)#

#~~0.33714+0.31032+0.11425#

#~~0.7617#

So the probability of one person "winning" is

Let

The probability of at least 3 of the friends winning is

#"P"(Z >= 3)#

#="P"(Z = 3) + "P"(Z = 4)#

#~~((4),(3))(0.7617)^3(0.2383)^1+((4),(4))(0.7617)^4(0.2383)^0#

#~~0.6319+0.3366#

#=0.9685#

So the probability of at least 3 out of 4 friends winning is