If a fair coin is tossed $10$ times. Find the probability of
a). Exactly $6$ heads
b). At least $6$ heads
c). At most $6$ heads
Answer
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Hint: We will first assume the number heads appearing as $x$. If the coin has a Bernoulli trial then $x$ must have a binomial distribution. From the distribution we will find the value of $\text{P}\left( \text{X}=x \right)$ by using the formula $\text{P}\left( \text{X}=x \right)={}^{n}{{\text{C}}_{x}}{{q}^{n-x}}{{p}^{x}}$. Here we will substitute the values of $n$ and $p$ from the given data and simplify the value of $\text{P}\left( \text{X}=x \right)$. Now to find the probabilities of each case we need to apply the values of $x$according to the condition, in $\text{P}\left( \text{X}=x \right)$. In the first condition we need to find the probability of getting $6$ heads, so we will substitute the value $x=6$ in $\text{P}\left( \text{X}=x \right)$ and we will find the value of probability. For the second case we need to find the probability of getting At least $6$ heads, so we will substitute $x\ge 6$ in $\text{P}\left( \text{X}=x \right)$ and we will find the value of probability. For the third case we need to find the probability of getting At most $6$ heads, so we will substitute $x\le 6$ in $\text{P}\left( \text{X}=x \right)$ and we will find the value of probability.
Complete step-by-step solution
Given that, A fair coin is tossed $10$ times, then
$n=10$
Let the number heads appear as $x$.
If coin toss is Bernoulli trial, then the binomial distribution for the variable $x$ I given by
$\text{P}\left( \text{X}=x \right)={}^{n}{{\text{C}}_{x}}{{q}^{n-x}}{{p}^{x}}$
Where
$p$ is the probability of getting head.
$q$ is the probability of not getting head.
We know that when we tossed a coin the probability of getting head is $\dfrac{1}{2}$, then
$p=\dfrac{1}{2}$
We know that sum of probabilities is equal to $1$, then
$p+q=1$
$\Rightarrow q=1-\dfrac{1}{2}=\dfrac{1}{2}$
Hence the distribution is
$\begin{align}
& \text{P}\left( \text{X}=x \right)={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10-x}}{{\left( \dfrac{1}{2} \right)}^{x}} \\
& ={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10-x+x}} \\
& ={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
\end{align}$
Now the probability of getting exactly $6$ heads is
$\text{P}\left( \text{X}=6 \right)={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}$
We know that ${}^{n}{{\text{C}}_{r}}=\dfrac{n!}{\left( n-r \right)!.r!}$, then
$\begin{align}
& \text{P}\left( \text{X}=6 \right)=\dfrac{10!}{\left( 10-6 \right)!6!}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{10\times 9\times 8\times 7\times 6!}{4!.6!}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{105}{512} \\
\end{align}$
Hence the probability of getting exactly $6$ heads is $\dfrac{105}{512}$.
Now the probability of getting at least $6$ heads is
$\begin{align}
& \text{P}\left( \text{X}\ge \text{6} \right)=\text{P}\left( \text{X}=\text{6} \right)+\text{P}\left( \text{X}=7 \right)+\text{P}\left( \text{X}=8 \right)+\text{P}\left( \text{X}=9 \right)+\text{P}\left( \text{X}=10 \right) \\
& ={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{6!\left( 10-6 \right)!}+\dfrac{10!}{7!\left( 10-7 \right)!}+\dfrac{10!}{8!\left( 10-8 \right)!}+\dfrac{10!}{9!\left( 10-9 \right)!}+\dfrac{10!}{10!\left( 10-10 \right)!} \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ 210+120+45+10+1 \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left( 386 \right) \\
& =\dfrac{193}{512} \\
\end{align}$
Hence the probability of getting at least $6$ heads is $\dfrac{193}{512}$
Now the probability of getting at most $6$ heads is
$\begin{align}
& \text{P}\left( \text{X}\le \text{6} \right)=\text{P}\left( \text{X}=\text{6} \right)+\text{P}\left( \text{X}=5 \right)+\text{P}\left( \text{X}=4 \right)+\text{P}\left( \text{X}=3 \right)+\text{P}\left( \text{X}=2 \right)+\text{P}\left( \text{X}=1 \right)+\text{P}\left( \text{X}=0 \right) \\
& ={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{5}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{4}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{3}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{2}}{{\left( \dfrac{1}{2} \right)}^{10}}+\,{}^{10}{{\text{C}}_{1}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{0}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{6!\left( 10-6 \right)!}+\dfrac{10!}{5!\left( 10-5 \right)!}+\dfrac{10!}{4!\left( 10-4 \right)!}+\dfrac{10!}{3!\left( 10-3 \right)!}+\dfrac{10!}{2!\left( 10-2 \right)!}+\dfrac{10!}{1!\left( 10-1 \right)!}+\dfrac{10!}{0!\left( 10-0 \right)!} \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ 210+252+210+120+45+10+1 \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left( 848 \right) \\
& =\dfrac{53}{64} \\
\end{align}$
Hence the probability of getting at most $6$ heads is $\dfrac{53}{64}$.
Note: Please note the difference between the terms ‘At least’ and ‘At most’. Mathematically ‘At least’ is similar to ‘Greater than or equal to’ and ‘At most’ is similar to ‘Less than or equal to’. Please note another point that $0!=1!=1$. We can also find the value of $\text{P}\left( \text{X}\le \text{6} \right)$ as
\[\begin{align}
& \text{P}\left( \text{X}\le \text{6} \right)=1-\text{P}\left( \text{X6} \right) \\
& =1-\text{P}\left( \text{X}=7 \right)-\text{P}\left( \text{X}=8 \right)-\text{P}\left( \text{X}=9 \right)-\text{P}\left( \text{X}=10 \right) \\
& =1-{}^{10}{{\text{C}}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{7!\left( 10-7 \right)!}+\dfrac{10!}{8!\left( 10-8 \right)!}+\dfrac{10!}{9!\left( 10-9 \right)!}+\dfrac{10!}{10!\left( 10-10 \right)!} \right] \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left( 120+45+10+1 \right) \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left( 176 \right) \\
& =1-\dfrac{11}{64} \\
& =\dfrac{53}{64} \\
\end{align}\]
From both the methods we get the same answer.
Complete step-by-step solution
Given that, A fair coin is tossed $10$ times, then
$n=10$
Let the number heads appear as $x$.
If coin toss is Bernoulli trial, then the binomial distribution for the variable $x$ I given by
$\text{P}\left( \text{X}=x \right)={}^{n}{{\text{C}}_{x}}{{q}^{n-x}}{{p}^{x}}$
Where
$p$ is the probability of getting head.
$q$ is the probability of not getting head.
We know that when we tossed a coin the probability of getting head is $\dfrac{1}{2}$, then
$p=\dfrac{1}{2}$
We know that sum of probabilities is equal to $1$, then
$p+q=1$
$\Rightarrow q=1-\dfrac{1}{2}=\dfrac{1}{2}$
Hence the distribution is
$\begin{align}
& \text{P}\left( \text{X}=x \right)={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10-x}}{{\left( \dfrac{1}{2} \right)}^{x}} \\
& ={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10-x+x}} \\
& ={}^{10}{{\text{C}}_{x}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
\end{align}$
Now the probability of getting exactly $6$ heads is
$\text{P}\left( \text{X}=6 \right)={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}$
We know that ${}^{n}{{\text{C}}_{r}}=\dfrac{n!}{\left( n-r \right)!.r!}$, then
$\begin{align}
& \text{P}\left( \text{X}=6 \right)=\dfrac{10!}{\left( 10-6 \right)!6!}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{10\times 9\times 8\times 7\times 6!}{4!.6!}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}.\dfrac{1}{{{2}^{10}}} \\
& =\dfrac{105}{512} \\
\end{align}$
Hence the probability of getting exactly $6$ heads is $\dfrac{105}{512}$.
Now the probability of getting at least $6$ heads is
$\begin{align}
& \text{P}\left( \text{X}\ge \text{6} \right)=\text{P}\left( \text{X}=\text{6} \right)+\text{P}\left( \text{X}=7 \right)+\text{P}\left( \text{X}=8 \right)+\text{P}\left( \text{X}=9 \right)+\text{P}\left( \text{X}=10 \right) \\
& ={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{6!\left( 10-6 \right)!}+\dfrac{10!}{7!\left( 10-7 \right)!}+\dfrac{10!}{8!\left( 10-8 \right)!}+\dfrac{10!}{9!\left( 10-9 \right)!}+\dfrac{10!}{10!\left( 10-10 \right)!} \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ 210+120+45+10+1 \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left( 386 \right) \\
& =\dfrac{193}{512} \\
\end{align}$
Hence the probability of getting at least $6$ heads is $\dfrac{193}{512}$
Now the probability of getting at most $6$ heads is
$\begin{align}
& \text{P}\left( \text{X}\le \text{6} \right)=\text{P}\left( \text{X}=\text{6} \right)+\text{P}\left( \text{X}=5 \right)+\text{P}\left( \text{X}=4 \right)+\text{P}\left( \text{X}=3 \right)+\text{P}\left( \text{X}=2 \right)+\text{P}\left( \text{X}=1 \right)+\text{P}\left( \text{X}=0 \right) \\
& ={}^{10}{{\text{C}}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{5}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{4}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{3}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{2}}{{\left( \dfrac{1}{2} \right)}^{10}}+\,{}^{10}{{\text{C}}_{1}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{\text{C}}_{0}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{6!\left( 10-6 \right)!}+\dfrac{10!}{5!\left( 10-5 \right)!}+\dfrac{10!}{4!\left( 10-4 \right)!}+\dfrac{10!}{3!\left( 10-3 \right)!}+\dfrac{10!}{2!\left( 10-2 \right)!}+\dfrac{10!}{1!\left( 10-1 \right)!}+\dfrac{10!}{0!\left( 10-0 \right)!} \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left[ 210+252+210+120+45+10+1 \right] \\
& ={{\left( \dfrac{1}{2} \right)}^{10}}\left( 848 \right) \\
& =\dfrac{53}{64} \\
\end{align}$
Hence the probability of getting at most $6$ heads is $\dfrac{53}{64}$.
Note: Please note the difference between the terms ‘At least’ and ‘At most’. Mathematically ‘At least’ is similar to ‘Greater than or equal to’ and ‘At most’ is similar to ‘Less than or equal to’. Please note another point that $0!=1!=1$. We can also find the value of $\text{P}\left( \text{X}\le \text{6} \right)$ as
\[\begin{align}
& \text{P}\left( \text{X}\le \text{6} \right)=1-\text{P}\left( \text{X6} \right) \\
& =1-\text{P}\left( \text{X}=7 \right)-\text{P}\left( \text{X}=8 \right)-\text{P}\left( \text{X}=9 \right)-\text{P}\left( \text{X}=10 \right) \\
& =1-{}^{10}{{\text{C}}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}-{}^{10}{{\text{C}}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left[ \dfrac{10!}{7!\left( 10-7 \right)!}+\dfrac{10!}{8!\left( 10-8 \right)!}+\dfrac{10!}{9!\left( 10-9 \right)!}+\dfrac{10!}{10!\left( 10-10 \right)!} \right] \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left( 120+45+10+1 \right) \\
& =1-{{\left( \dfrac{1}{2} \right)}^{10}}\left( 176 \right) \\
& =1-\dfrac{11}{64} \\
& =\dfrac{53}{64} \\
\end{align}\]
From both the methods we get the same answer.
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