Let "X=" the number of male children: "X\\sim Bin(n, p)"
"P(X=x)=\\dbinom{n}{k}p^x(1-p)^{n-x}" Given "p=0.6, n=6."
a)
"P(X=3)=\\dbinom{6}{3}0.6^3(1-0.6)^{6-3}=0.27648" b)
"P(X\\geq2)=1-P(X=0)-P(X=1)=""=1-\\dbinom{6}{0}0.6^0(1-0.6)^{6-0}-\\dbinom{6}{1}0.6^1(1-0.6)^{6-1}=""=0.95904"
c)
"P(X=6)=\\dbinom{6}{6}0.6^6(1-0.6)^{6-6}=0.046656"
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