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consider tossing a biased coin whose probability is of coming up heads is 0.3. if 40 trials are performed, what values of k would have probabilities closest to half the probability of the expected value of k

Guest Apr 2, 2015

Best Answer 

 #1
avatar+94105 
+10

consider tossing a biased coin whose probability is of coming up heads is 0.3. if 40 trials are performed, what values of k would have probabilities closest to half the probability of the expected value of k

By the expected value of k, I am going to assume that you mean the expected probability IF the the coin was not biased.

 

$$\begin{array}{rll}
2\times \binom{40}{k}(0.3)^k(0.7)^{(40-k)}&=&\binom{40}{k}(0.5)^k(0.5)^{(40-k)}\\\\
2\times (0.3)^k(0.7)^{(40-k)}&=&(0.5)^{40}\\\\
2\times \left(\frac{3}{10}\right)^k\left(\frac{7}{10}\right)^{(40-k)}&=&\left(\frac{1}{2}\right)^{40}\\\\
2\times \frac{3^k\times 7^{(40-k)}}{10^k\times {10^{(40-k)}}}&=&\frac{1}{2^{40}}\\\\
\frac{3^k\times 7^{40}}{10^{40}\times 7^k}&=&\frac{1}{2^{41}}\\\\
\frac{3^k}{ 7^k}&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
\left(\frac{3}{ 7}\right)^k&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
log\left[\left(\frac{3}{ 7}\right)^k\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k\times log\left[\left(\frac{3}{ 7}\right)\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k &=&\frac{log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]}{log\left(\frac{3}{ 7}\right)}\\\\



\end{array}$$

 

$$\left({\frac{{log}_{10}\left({\frac{{{\mathtt{10}}}^{{\mathtt{40}}}}{\left({{\mathtt{2}}}^{{\mathtt{41}}}{\mathtt{\,\times\,}}{{\mathtt{7}}}^{{\mathtt{40}}}\right)}}\right)}{{log}_{10}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}}\right) = {\mathtt{16.702\: \!552\: \!085\: \!923\: \!092\: \!1}}$$

 

The closest one is k=17

Melody  Apr 2, 2015
 #1
avatar+94105 
+10
Best Answer

consider tossing a biased coin whose probability is of coming up heads is 0.3. if 40 trials are performed, what values of k would have probabilities closest to half the probability of the expected value of k

By the expected value of k, I am going to assume that you mean the expected probability IF the the coin was not biased.

 

$$\begin{array}{rll}
2\times \binom{40}{k}(0.3)^k(0.7)^{(40-k)}&=&\binom{40}{k}(0.5)^k(0.5)^{(40-k)}\\\\
2\times (0.3)^k(0.7)^{(40-k)}&=&(0.5)^{40}\\\\
2\times \left(\frac{3}{10}\right)^k\left(\frac{7}{10}\right)^{(40-k)}&=&\left(\frac{1}{2}\right)^{40}\\\\
2\times \frac{3^k\times 7^{(40-k)}}{10^k\times {10^{(40-k)}}}&=&\frac{1}{2^{40}}\\\\
\frac{3^k\times 7^{40}}{10^{40}\times 7^k}&=&\frac{1}{2^{41}}\\\\
\frac{3^k}{ 7^k}&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
\left(\frac{3}{ 7}\right)^k&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
log\left[\left(\frac{3}{ 7}\right)^k\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k\times log\left[\left(\frac{3}{ 7}\right)\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k &=&\frac{log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]}{log\left(\frac{3}{ 7}\right)}\\\\



\end{array}$$

 

$$\left({\frac{{log}_{10}\left({\frac{{{\mathtt{10}}}^{{\mathtt{40}}}}{\left({{\mathtt{2}}}^{{\mathtt{41}}}{\mathtt{\,\times\,}}{{\mathtt{7}}}^{{\mathtt{40}}}\right)}}\right)}{{log}_{10}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}}\right) = {\mathtt{16.702\: \!552\: \!085\: \!923\: \!092\: \!1}}$$

 

The closest one is k=17

Melody  Apr 2, 2015
 #2
avatar+92673 
0

Hey, Melody....a brief question......what was the motivation for multiplying by "2" on the left hand side in the top equation ???

Go slow with that explanation......you know me and "counting" problems....LOL!!!!

 

  

CPhill  Apr 2, 2015
 #3
avatar+92673 
+5

Nevermind.....I see  it.....(1/2) of the expected probability of a "normal" coin = 2 x the probability of our "biased" coin....

 BTW....nice answer....!!!

  

CPhill  Apr 2, 2015
 #4
avatar+94105 
0

Thanks Chris.  :)

Melody  Apr 3, 2015

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