The probability of getting a boy is `1.09/(1.09+1.00)=0.5215`

Let `X =` number of boys in the family.

Here,

`n = 6`,

`p = 0.5215`,

`q = 1 − 0.52153 = 0.4785`

When `x=3`:

` P(X)` `=C_x^np^xq^(n-x)` `=C_3^6(0.5215)^3(0.4785)^3` `=0.31077`

When `x=4`:

` P(X)` `=C_4^6(0.5215)^4(0.4785)^2` `=0.25402`

When `x=5`:

`P(X)` `=C_5^6(0.5215)^5(0.4785)^1` `=0.11074`

When `x=6`:

`P(X)` `=C_6^6(0.5215)^6(0.4785)^0` `=2.0115xx10^-2`

So the probability of getting at least 3 boys is:

`"Probability"=P(X>=3)`

`=0.31077+0.25402+` `0.11074+` `2.0115xx10^-2`

`=0.69565`

**NOTE:** We could have calculated it like this:

`P(X>=3)` `=1-(P(x_0)+P(x_1)+P(x_2))`

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