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
+118609 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
+118609 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
+128475 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!!!!
