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One enterprising gambler decides to take a bet that if he flips a coin 6 times in a row and then rolls a die 6 times in a row, he will get exactly 3 heads or 3 sixes. What are his odds of winning the bet? (Answer to the nearest hundredth percent)

Guest Mar 30, 2015

Best Answer 

 #1
avatar+91469 
+5

$$P= \binom{6}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3\times \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$

 

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{{\mathtt{3}}} = {\mathtt{0.016\: \!744\: \!898\: \!834\: \!019\: \!2}}$$

 

= 1.7644...%

 

chance of winning =1.76% to the nearest 100th of a percent.

 

NOTE:

$$\binom{6}{3}\quad$is the same as $ ^6C_3 \\\\
$Sometimes when I am being a little lazy I may also write it as 6C3.$$$

Melody  Mar 31, 2015
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1+0 Answers

 #1
avatar+91469 
+5
Best Answer

$$P= \binom{6}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3\times \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$

 

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{{\mathtt{3}}} = {\mathtt{0.016\: \!744\: \!898\: \!834\: \!019\: \!2}}$$

 

= 1.7644...%

 

chance of winning =1.76% to the nearest 100th of a percent.

 

NOTE:

$$\binom{6}{3}\quad$is the same as $ ^6C_3 \\\\
$Sometimes when I am being a little lazy I may also write it as 6C3.$$$

Melody  Mar 31, 2015

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