Answer

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Hint: Using the binomial expansion theorem we need to find a probability that out of six cases at least five of them are solved.

It is given that $25%$ of the new cases of child labour are reported in the police station are solved.

So, the probability of cases being solved is,

$\begin{align}

& =\dfrac{25}{100} \\

& =\dfrac{1}{4} \\

\end{align}$

Let the probability of cases getting solved be denoted by $'P'$. So, we have

$P=\dfrac{1}{4}$

Then the probabilities of cases not getting solved will be,

$Q=\dfrac{3}{4}$

Now as it is said six new cases are reported.

Now we need to find the probability that out of six new cases at least five of them are solved or six cases are solved.

Here we will apply binomial expansion formula.

As per binomial expansion, we choose $'k'$ numbers out of $'n'$, that is in how many ways we can choose $'k'$elements from a set of $'n'$.

So, the general formula would be

\[{{(a+b)}^{n}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}

n \\

k \\

\end{matrix} \right){{a}^{n-k}}{{b}^{k}}}\]

Where ‘a’ and ‘b’ are two events.

So, applying this formula in our question, we have

Probability that at least five cases are solved be,

$R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{Q}^{6-k}}{{P}^{k}}}$

Expanding this, we get

$R\left( X\ge 5 \right)=\left( \begin{matrix}

6 \\

5 \\

\end{matrix} \right){{Q}^{6-5}}{{P}^{5}}+\left( \begin{matrix}

6 \\

6 \\

\end{matrix} \right){{Q}^{6-6}}{{P}^{6}}$

Substituting the values, we get

\[\begin{align}

& R\left( X\ge 5 \right)=\dfrac{6!}{5!(6-5)!}{{\left( \dfrac{3}{4} \right)}^{1}}{{\left( \dfrac{1}{4} \right)}^{5}}+\dfrac{6!}{6!(6-6)!}{{\left( \dfrac{3}{4} \right)}^{0}}{{\left( \dfrac{1}{4} \right)}^{6}} \\

& \Rightarrow R\left( X\ge 5 \right)=6\left( \dfrac{3}{4} \right)\left( \dfrac{1}{1024} \right)+(1)(1)\left( \dfrac{1}{4096} \right) \\

& \Rightarrow R\left( X\ge 5 \right)=\left( \dfrac{18}{4096} \right)+\left( \dfrac{1}{4096} \right) \\

& \Rightarrow R\left( X\ge 5 \right)=\left( \dfrac{19}{4096} \right) \\

\end{align}\]

So, the probability that at least 5 of the new cases are solve will be \[\left( \dfrac{19}{4096} \right)\].

Note: Student gets confused between $P=\dfrac{1}{4}$ and $Q=\dfrac{3}{4}$. In the binomial expansion if they substitute P as Q and Q as P, i.e.,

$R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{Q}^{6-k}}{{P}^{k}}}$is written as $R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{P}^{6-k}}{{Q}^{k}}}$, then they will get the wrong answer.

It is given that $25%$ of the new cases of child labour are reported in the police station are solved.

So, the probability of cases being solved is,

$\begin{align}

& =\dfrac{25}{100} \\

& =\dfrac{1}{4} \\

\end{align}$

Let the probability of cases getting solved be denoted by $'P'$. So, we have

$P=\dfrac{1}{4}$

Then the probabilities of cases not getting solved will be,

$Q=\dfrac{3}{4}$

Now as it is said six new cases are reported.

Now we need to find the probability that out of six new cases at least five of them are solved or six cases are solved.

Here we will apply binomial expansion formula.

As per binomial expansion, we choose $'k'$ numbers out of $'n'$, that is in how many ways we can choose $'k'$elements from a set of $'n'$.

So, the general formula would be

\[{{(a+b)}^{n}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}

n \\

k \\

\end{matrix} \right){{a}^{n-k}}{{b}^{k}}}\]

Where ‘a’ and ‘b’ are two events.

So, applying this formula in our question, we have

Probability that at least five cases are solved be,

$R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{Q}^{6-k}}{{P}^{k}}}$

Expanding this, we get

$R\left( X\ge 5 \right)=\left( \begin{matrix}

6 \\

5 \\

\end{matrix} \right){{Q}^{6-5}}{{P}^{5}}+\left( \begin{matrix}

6 \\

6 \\

\end{matrix} \right){{Q}^{6-6}}{{P}^{6}}$

Substituting the values, we get

\[\begin{align}

& R\left( X\ge 5 \right)=\dfrac{6!}{5!(6-5)!}{{\left( \dfrac{3}{4} \right)}^{1}}{{\left( \dfrac{1}{4} \right)}^{5}}+\dfrac{6!}{6!(6-6)!}{{\left( \dfrac{3}{4} \right)}^{0}}{{\left( \dfrac{1}{4} \right)}^{6}} \\

& \Rightarrow R\left( X\ge 5 \right)=6\left( \dfrac{3}{4} \right)\left( \dfrac{1}{1024} \right)+(1)(1)\left( \dfrac{1}{4096} \right) \\

& \Rightarrow R\left( X\ge 5 \right)=\left( \dfrac{18}{4096} \right)+\left( \dfrac{1}{4096} \right) \\

& \Rightarrow R\left( X\ge 5 \right)=\left( \dfrac{19}{4096} \right) \\

\end{align}\]

So, the probability that at least 5 of the new cases are solve will be \[\left( \dfrac{19}{4096} \right)\].

Note: Student gets confused between $P=\dfrac{1}{4}$ and $Q=\dfrac{3}{4}$. In the binomial expansion if they substitute P as Q and Q as P, i.e.,

$R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{Q}^{6-k}}{{P}^{k}}}$is written as $R\left( X\ge 5 \right)=\sum\limits_{k=5}^{6}{\left( \begin{matrix}

6 \\

k \\

\end{matrix} \right){{P}^{6-k}}{{Q}^{k}}}$, then they will get the wrong answer.

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