Question

If 4- digits number greater that 5000 are randomly formed the digits 0,1,3,5 and 7, then what is the probability of forming a number divisible by 5, when

(i) the digits may be repeated?

(ii) the repetition of digits in not allowed?

Or

A card is drawn from a deck of 52 cards . Find the probability of getting

(i) an ace.

(ii) a king or a heart or a red card.

Answer 1

(i) let 4-digits number be formed as

I II III IV

$I\to$ first place can be filled in two ways. using digit 5 or 7 .[$\because$ number greater than 5000]

Each of the remaining three places can be filled in 5 ways.

$\therefore$ Total numbers so formed $=2\times 5\times 5\times 5=250$

If the number is divisible by 5. then IV place can be filled in 2 ways using digit 0 or 5.

$\therefore$ Number, so formed that are divisible by 5

$2\times 5\times 5\times 2=100$

Hence, required probability =$\frac{Favourablecases}{Totalcases}$

$=\frac{100}{250}=\frac{2}{5}$

(ii) Let 4 places in a 4-digits number be formed as I II III IV

I place can be filled in two ways, using digit 5 or 7.

Then places II , III,IV may be filled,

$=4\times 3\times 2$

$\therefore$Total number of are so formed $=2\times 4\times 3\times 2$

$\Rightarrow n\left(S\right)=48$ways.

If the number is divisible by 5, o or 5 is placed at unit place, ie. at IV place.

$\begin{array}{cccc}I& II& III& IV\\ 5& -& -& 0\end{array}$

$\begin{array}{cccc}I& II& III& IV\\ 5& -& -& 0\end{array}$

$\begin{array}{cccc}I& II& III& IV\\ 7& -& -& 5\end{array}$

In each case, places II and III can be filed in $3\times 2.$i.e, 6 ways.

$\therefore n\left(E\right)=6+6+6=18$

Hence , required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{18}{48}=\frac{3}{8}$

Or

(i) There are four aces in a pack of 52 cards, out of which one ace card can be drawn in ${}^4{C}_1$ways.

$\therefore$Favorable number of outcomes =${}^4{C}_1=4$

So, required probability=$\frac{4}{52}=\frac{1}{13}$

(ii) There are 26 red card,out of which one red card can be drawn in ${}^{26}{C}_1=26$

$\therefore$ Favorable number of number of outcomes =${}^{26}{C}_1=26$

So, required probability $=\frac{26}{52}=\frac{1}{2}$

(iii) There are 26 red cards including 13 hearts plus 2 red king and there are 2 more kings,. Therefore there are 28 ways cards which are either red or king or heart , out of which one card can be drawn in ${}^{26}{C}_1=26$ ways,

$\therefore$ Favorable number of outcomes =28

Each of 52 cards, one card can be drawn in ${}^{52}{C}_1=52$ ways.

$\therefore$ Total number of outcomes =${}^{52}{C}_1=52$

So, required probability $=\frac{28}{52}=\frac{7}{13}$