Question

The probability that a doctor successfully performs an operation is $80\%$ . What is the value of $625$ times the probability that at least $3$ operations out of $4$ conducted by him will be successful?

Answer 1

We will find the probability that at least $3$ operations out of $4$ conducted by him will be successful by using Binomial distribution. The formula for binomial distribution is $P(r:n,p) = {}^n{C_r}{p^r}{(q)^{n - r}}$ where $n$ = the number of experiments, $r = 0,1,2,..$, $p$ = Probability of Success in a single experiment, $q$ = Probability of Failure in a single experiment = $1 - p$.

Complete answer: Let the probability of successful operation be $p$ = $80\%$ = $\dfrac{{80}}{{100}}$ = $\dfrac{4}{5}$
And the probability of an unsuccessful operation be $q$ = $1 - p$ = $1 - \dfrac{4}{5}$ = $\dfrac{1}{5}$
The total number of operations conducted by doctor is $n$ = $4$
The probability of $x$ successful operations is given by $P(X = r) = {}^n{C_r}{p^r}{(q)^{n - r}} = {}^4{C_r}{\left( {\dfrac{4}{5}} \right)^r}{\left( {\dfrac{1}{5}} \right)^{4 - r}}$
We need to find the probability of at least $3$ operations out of $4$ conducted by him will be successful, which is given by
$P(X \geqslant 3) = P(X = 3) + P(X = 4)$
Now,
$P(X = 3) = {}^4{C_3}{\left( {\dfrac{4}{5}} \right)^3}{\left( {\dfrac{1}{5}} \right)^{4 - 3}} = {}^4{C_3}{\left( {\dfrac{4}{5}} \right)^3}{\left( {\dfrac{1}{5}} \right)^1}$
$P(X = 4) = {}^4{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^{4 - 4}} = {}^4{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^0}$
Substituting these values in $P(X \geqslant 3) = P(X = 3) + P(X = 4)$,
$P(X \geqslant 3) = {}^4{C_3}{\left( {\dfrac{4}{5}} \right)^3}{\left( {\dfrac{1}{5}} \right)^1} + {}^4{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^0}$
The combination formula is given by ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Applying the combination formula in above equation,
$P(X \geqslant 3) = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}{\left( {\dfrac{4}{5}} \right)^3}{\left( {\dfrac{1}{5}} \right)^1} + \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^0}$
Simplifying the factorial part,
$P(X \geqslant 3) = \dfrac{{4!}}{{3!1!}}{\left( {\dfrac{4}{5}} \right)^3}{\left( {\dfrac{1}{5}} \right)^1} + \dfrac{{4!}}{{4!0!}}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^0}$
We know, ${\left( {\dfrac{1}{5}} \right)^0} = 1$ and ${\left( {\dfrac{1}{5}} \right)^1} = \left( {\dfrac{1}{5}} \right)$,
$\therefore P(X \geqslant 3) = \dfrac{{4!}}{{3!1!}}{\left( {\dfrac{4}{5}} \right)^3}\left( {\dfrac{1}{5}} \right) + \dfrac{{4!}}{{4!0!}}{\left( {\dfrac{4}{5}} \right)^4}$
Simplifying the factorial part,
$\therefore P(X \geqslant 3) = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 1}}{\left( {\dfrac{4}{5}} \right)^3}\left( {\dfrac{1}{5}} \right) + \dfrac{{4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}}{\left( {\dfrac{4}{5}} \right)^4}$
Cancelling factorial terms,
$P(X \geqslant 3) = 4{\left( {\dfrac{4}{5}} \right)^3}\left( {\dfrac{1}{5}} \right) + {\left( {\dfrac{4}{5}} \right)^4}$
Rearranging the terms,
$P(X \geqslant 3) = {\left( {\dfrac{4}{5}} \right)^3}\left( {\dfrac{4}{5}} \right) + {\left( {\dfrac{4}{5}} \right)^4}$
Applying the form, ${a^m}.{a^n} = {a^{m + n}}$ in the above equation, we get,
$P(X \geqslant 3) = {\left( {\dfrac{4}{5}} \right)^{3 + 1}} + {\left( {\dfrac{4}{5}} \right)^4}$
Adding the powers,
$P(X \geqslant 3) = {\left( {\dfrac{4}{5}} \right)^4} + {\left( {\dfrac{4}{5}} \right)^4}$
We can write this equation as,
$P(X \geqslant 3) = \dfrac{{{{\left( 4 \right)}^4}}}{{{{\left( 5 \right)}^4}}} + \dfrac{{{{\left( 4 \right)}^4}}}{{{{\left( 5 \right)}^4}}}$
As the denominators are same, we can write,
$P(X \geqslant 3) = \dfrac{{{{\left( 4 \right)}^4} + {{\left( 4 \right)}^4}}}{{{{\left( 5 \right)}^4}}}$
Taking ${4^{th}}$ power of $4$ and $5$ ,
$P(X \geqslant 3) = \dfrac{{256 + 256}}{{625}}$
Adding the terms,
$P(X \geqslant 3) = \dfrac{{512}}{{625}}$
The value of $625$ times the probability that at least $3$ operations out of $4$ conducted by him will be successful is $P(X \geqslant 3) \times 625$ i.e.,
$P(X \geqslant 3) = \dfrac{{512}}{{625}} \times 625$
Cancelling the terms,
$P(X \geqslant 3) = 512$
The value of $625$ times the probability that at least $3$ operations out of $4$ conducted by him will be successful is $512$ .

Note:
Binomial distribution gives only two outcomes either success or failure. In binomial distribution, every trial is an independent trial i.eThe outcome of one trial does not depend on other trials and the probability of failure varies for each trial.