Solution
- This is a Binomial theorem question. The following are the parameters given:
[tex]\begin{gathered} n=10 \\ r=2 \\ p=0.6 \\ q=1-p=0.4 \\ \\ \text{ The formula to use is:} \\ P(r)=\sum_r^nnC_rp^rq^{n-r} \end{gathered}[/tex]- Thus, we have:
[tex]\begin{gathered} P(r\le2)=P(0)+P(1)+P(2) \\ P(0)=^{10}C_00.6^0\times0.4^{10}=0.0001048576 \\ P(1)=^{10}C_10.6^1\times0.4^9=0.001572864 \\ P(2)=^{10}C_20.6^2\times0.4^8=0.010616832 \\ \\ \therefore P(r\le2)=0.0001048576+0.001572864+0.010616832 \\ \\ P(r\le2)=0.0122945536\approx0.0123 \end{gathered}[/tex]
