Question

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?

Answer 1

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

`"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 `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

`"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 `~~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

`"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 `~~0.9685.`