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Hint: In each part, the number of favourable outcomes can be calculated manually. There is no need of applying the concept of permutation and combination.

Before proceeding with the question, we must know the definition of the probability. The probability of an event $X$ is defined as the ratio of the number of outcomes favourable to event $X$ and the total number of possible outcomes in the sample space. Mathematically,’

$P\left( X \right)=\dfrac{n\left( X \right)}{n\left( S \right)}...............\left( 1 \right)$

In this question, we have cards numbered from $11$ to $60$. So, the sample space $S$ is,

$S=\left\{ 11,12,13,14,................,58,59,60 \right\}$

So, total number of possible outcomes $n\left( S \right)$ which is same for each part in the question and is equal to,

$n\left( S \right)=50..............\left( 2 \right)$

(i) It is given that the card drawn is an odd number i.e. set $X$ is,

$X=\left\{ 11,13,15,.........,53,57,59 \right\}$

So, $n\left( X \right)=25.........\left( 3 \right)$

Substituting $n\left( X \right)=25$ from equation $\left( 3 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{25}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{1}{2} \\

\end{align}$

(ii) It is given that the card drawn is a perfect square number i.e. set $X$ is,

$X=\left\{ 16,25,36,49 \right\}$

So, $n\left( X \right)=4.........\left( 4 \right)$

Substituting $n\left( X \right)=4$ from equation $\left( 4 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{4}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{2}{25} \\

\end{align}$

(iii) It is given that the card drawn is a number which is divisible by $5$ i.e. set $X$ is,

$X=\left\{ 15,20,25,30,35,40,45,50,55,60 \right\}$

So, $n\left( X \right)=10.........\left( 5 \right)$

Substituting $n\left( X \right)=10$ from equation $\left( 5 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{10}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{1}{5} \\

\end{align}$

(iv) It is given that the card drawn is a prime number less than $20$ i.e. set $X$ is,

$X=\left\{ 11,13,17,19 \right\}$

So, $n\left( X \right)=4.........\left( 6 \right)$

Substituting $n\left( X \right)=4$ from equation $\left( 6 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{4}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{2}{25} \\

\end{align}$

Hence, the answer of $\left( i \right)$ is $\dfrac{1}{2}$, $\left( ii \right)$ is $\dfrac{2}{25}$, $\left( iii \right)$ is $\dfrac{1}{5}$, $\left( iv \right)$ is $\dfrac{2}{25}$.

Note: There is a possibility of committing mistakes while calculating the number of favourable outcomes. Since we are calculating the number of favourable outcomes by finding out the set $X$, there is a possibility that one may find out the wrong number.

Before proceeding with the question, we must know the definition of the probability. The probability of an event $X$ is defined as the ratio of the number of outcomes favourable to event $X$ and the total number of possible outcomes in the sample space. Mathematically,’

$P\left( X \right)=\dfrac{n\left( X \right)}{n\left( S \right)}...............\left( 1 \right)$

In this question, we have cards numbered from $11$ to $60$. So, the sample space $S$ is,

$S=\left\{ 11,12,13,14,................,58,59,60 \right\}$

So, total number of possible outcomes $n\left( S \right)$ which is same for each part in the question and is equal to,

$n\left( S \right)=50..............\left( 2 \right)$

(i) It is given that the card drawn is an odd number i.e. set $X$ is,

$X=\left\{ 11,13,15,.........,53,57,59 \right\}$

So, $n\left( X \right)=25.........\left( 3 \right)$

Substituting $n\left( X \right)=25$ from equation $\left( 3 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{25}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{1}{2} \\

\end{align}$

(ii) It is given that the card drawn is a perfect square number i.e. set $X$ is,

$X=\left\{ 16,25,36,49 \right\}$

So, $n\left( X \right)=4.........\left( 4 \right)$

Substituting $n\left( X \right)=4$ from equation $\left( 4 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{4}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{2}{25} \\

\end{align}$

(iii) It is given that the card drawn is a number which is divisible by $5$ i.e. set $X$ is,

$X=\left\{ 15,20,25,30,35,40,45,50,55,60 \right\}$

So, $n\left( X \right)=10.........\left( 5 \right)$

Substituting $n\left( X \right)=10$ from equation $\left( 5 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{10}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{1}{5} \\

\end{align}$

(iv) It is given that the card drawn is a prime number less than $20$ i.e. set $X$ is,

$X=\left\{ 11,13,17,19 \right\}$

So, $n\left( X \right)=4.........\left( 6 \right)$

Substituting $n\left( X \right)=4$ from equation $\left( 6 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( X \right)=\dfrac{4}{50} \\

& \Rightarrow P\left( X \right)=\dfrac{2}{25} \\

\end{align}$

Hence, the answer of $\left( i \right)$ is $\dfrac{1}{2}$, $\left( ii \right)$ is $\dfrac{2}{25}$, $\left( iii \right)$ is $\dfrac{1}{5}$, $\left( iv \right)$ is $\dfrac{2}{25}$.

Note: There is a possibility of committing mistakes while calculating the number of favourable outcomes. Since we are calculating the number of favourable outcomes by finding out the set $X$, there is a possibility that one may find out the wrong number.

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