Colby is practicing his basketball shots. He knows that he has a 0.7 chance of making
each 2-point shot and a 0.4 chance of making each 3-point shot. If he takes five 2-point
and five 3-point shots, what is his probability of earning 20 points or more? Express your
answer as a decimal to the nearest thousandth.
+118724 This is a messy one 
P(3)=0.4 P(not3)=0.6
seperately
P(2)=0.7 P(not2)=0.3
Now lets see what will work
5(3s)+5(2s)=15+10=25 points
5(3s)+4(2s)=15+8=23 points
5(3s)+3(2s)=15+6=21 points
5(3s)+2(2s)=15+4=19 points TOO FEW
4(3s)+5(2s)=12+10=22 points
4(3s)+4(2s)=12+8=20 points
3(3s)+5(2s)=9+10=19 points TOO FEW
---------------
P(3)=0.4 P(not3)=0.6 seperately P(2)=0.7 P(not2)=0.3
P[5(3s)+5(2s)]= $${{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}$$
P[5(3s)+4(2s)]
=$$0.4^5 * 5C4*0.7^4*0.3^1$$ $$=0.4^5 * 5* 0.7^4*0.3=0.4^5*1.5*0.7^4$$
P[5(3s)+3(2s)]
=$$0.4^5* 5C3*0.7^3*0.3^2=0.4^5*10*0.7^3*0.3^2$$
P[4(3s)+5(2s)]
$$=5C4*0.4^4*0.6\;\;*\;\;0.7^5$$
P[4(3s)+4(2s)]
$$=5C4*0.4^4*0.6\;\;*\;\;5C4*0.7^4*0.3=5*0.4^4*0.6\;\;*\;\;5*0.7^4*0.3$$
Now add all the separate probablilies together for a final answer.
$${{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\mathtt{1.5}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.6}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.3}} = {\mathtt{0.051\: \!595\: \!980\: \!8}}$$
Now if I did not make any careless mistakes that should be the answer.
+118724 This is a messy one
P(3)=0.4 P(not3)=0.6
seperately
P(2)=0.7 P(not2)=0.3
Now lets see what will work
5(3s)+5(2s)=15+10=25 points
5(3s)+4(2s)=15+8=23 points
5(3s)+3(2s)=15+6=21 points
5(3s)+2(2s)=15+4=19 points TOO FEW
4(3s)+5(2s)=12+10=22 points
4(3s)+4(2s)=12+8=20 points
3(3s)+5(2s)=9+10=19 points TOO FEW
---------------
P(3)=0.4 P(not3)=0.6 seperately P(2)=0.7 P(not2)=0.3
P[5(3s)+5(2s)]= $${{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}$$
P[5(3s)+4(2s)]
=$$0.4^5 * 5C4*0.7^4*0.3^1$$ $$=0.4^5 * 5* 0.7^4*0.3=0.4^5*1.5*0.7^4$$
P[5(3s)+3(2s)]
=$$0.4^5* 5C3*0.7^3*0.3^2=0.4^5*10*0.7^3*0.3^2$$
P[4(3s)+5(2s)]
$$=5C4*0.4^4*0.6\;\;*\;\;0.7^5$$
P[4(3s)+4(2s)]
$$=5C4*0.4^4*0.6\;\;*\;\;5C4*0.7^4*0.3=5*0.4^4*0.6\;\;*\;\;5*0.7^4*0.3$$
Now add all the separate probablilies together for a final answer.
$${{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\mathtt{1.5}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.6}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{0.7}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{0.3}} = {\mathtt{0.051\: \!595\: \!980\: \!8}}$$
Now if I did not make any careless mistakes that should be the answer.
