a) To verify that the Rams' computer system manager statement

sibuzwaW 2021-10-02 Answered
a) To verify that the Rams' computer system manager statement that the odds against losing straight tosses are 2047 to 1 is correct or not.
b) To find the probability that the Rams lost the call for the next two games, for a total of 13 straight losses.

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Expert Answer

ensojadasH
Answered 2021-10-03 Author has 6526 answers
a) The binomial probability is defined as,
\(\displaystyle{P}{\left({X}={k}\right)}={{C}_{{{k}}}^{{{n}}}}\cdot{p}^{{{k}}}\cdot{q}^{{{n}-{k}}}={\frac{{{n}!}}{{{k}!{\left({n}-{k}\right)}!}}}\times{p}^{{{k}}}\times{\left({1}-{p}\right)}^{{{n}-{k}}}\)
Where \(\displaystyle{n}=\) number of trials \(\displaystyle={11}\)
\(\displaystyle{p}=\) Probability of success \(\displaystyle={\frac{{{1}}}{{{2}}}}={0.5}\)
(Assuming heads and tails are equally likely)
Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use binomial distribution.
Now, we will find the binomial probability at \(\displaystyle{k}={11}\),
\(\displaystyle{P}{\left({X}={11}\right)}={{C}_{{{11}}}^{{{11}}}}\cdot{\left({0.5}\right)}^{{{11}}}\cdot{\left({1}-{0.5}\right)}^{{{11}-{11}}}\)
\(\displaystyle={\frac{{{11}!}}{{{11}!{\left({11}-{11}\right)}!}}}\times{\left({0.5}\right)}^{{{11}}}\times{\left({1}-{0.5}\right)}^{{{0}}}\)
\(\displaystyle={0.5}^{{{11}}}\)
\(\displaystyle={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{11}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}^{{{11}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2048}}}}\)
Thus we note that 1 out of 2048 outcomes result in losing eleven straight tosses, which implies that 2047 out of 2048 outcomes result in not losing eleven straight tosses and thus the ods against losing eleven straight tosses in 2047 to 1.
b) The binomial probability is defined as,
\(\displaystyle{P}{\left({X}={k}\right)}={{C}_{{{k}}}^{{{n}}}}\cdot{p}^{{{k}}}\cdot{q}^{{{n}-{k}}}={\frac{{{n}!}}{{{k}!{\left({n}-{k}\right)}!}}}\times{p}^{{{k}}}\times{\left({1}-{p}\right)}^{{{n}-{k}}}\)
Where \(\displaystyle{n}=\) number of trials \(\displaystyle={13}\)
\(\displaystyle{p}=\) Probability of success \(\displaystyle={\frac{{{1}}}{{{2}}}}={0.5}\)
(Assuming heads and tails are equally likely)
Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use binomial distribution.
Now, we will find the binomial probability at \(\displaystyle{k}={13}\)
\(\displaystyle{P}{\left({X}={13}\right)}={{C}_{{{13}}}^{{{13}}}}\cdot{\left({0.5}\right)}^{{{13}}}\cdot{\left({1}-{0.5}\right)}^{{{13}-{13}}}\)
\(\displaystyle={\frac{{{13}!}}{{{13}!{\left({13}-{13}\right)}!}}}\times{\left({0.5}\right)}^{{{13}}}\times{\left({1}-{0.5}\right)}^{{{0}}}\)
\(\displaystyle={0.5}^{{{13}}}\)
\(\displaystyle={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{13}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}^{{{13}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{8192}}}}\approx{0.0001}\)
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