Question

From a pack of 30 cards marked with numbers 1,2,3.....30 .All cards whose numbers are multiples of 3 are removed. A card is now drawn at random. Find the probability of getting a card which is: (4 M) i. A prime number less than 10. ii. A number divisible by 5.​

Answer 1

GIVEN :-

• 30 cards numbered 1 - 30

• All cards whose numbers are multiples of 3 are removed.

TO FIND :-

• Find the probability of getting a card which is:

i. A prime number less than 10.

ii. A number divisible by 5.

iii. A perfect cube.

iv. A number that is a multiple of 6.

SOLUTION :-

SO THERE ARE 30 cards ( numbered 1 - 30 )

multiple of 3 between 1 - 30 are :-

3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30

so these cards have been removed

cards left in deck :- 1 , 2 , 4 , 5 , 7 , 8 , 10 , 11 , 13 ,

14 , 16 , 17 , 19 , 20 , 22 , 23 ,

25 , 26 , 28 , 29 ( total - 20 )

NOW ATQ :-

(i)

PRIME NUMBER LESS THAN 10 = 2 , 3 , 5 , 7

(but 3 is removed so we will take only 2 ,5 ,7 )

HENCE,

NUMBER OF CARDS IN PRIME NUMBER LESS THAN 10 IN CARD DECK = 2 , 5 , 7 (3 cards)

NOW TOTAL NUMBER OF CARDS = 20

$\implies \boxed{ \rm{ probability \: = \dfrac{no \: of \: fav \: events}{total \: no \: of \: events} }}$

$\implies \boxed { \boxed{ \rm{probability \: = \: \: \dfrac{3}{20} }}}$

(ii)

NUMBERS DIVISBLE BY 5 BELOW 30 ARE :-

5 , 10 , 15 , 20 , 25 , 30

( but 15 and 30 are removed )

HENCE ,

NUMBER OF CARD DIVISBLE BY 5 IN DECK :-

NUMBER OF CARD DIVISBLE BY 5 IN DECK :-5 , 10 , 20 , 25 ( TOTAL 4 CARDS )

NOW TOTAL NO OF CARDS = 20

$\implies \boxed{ \rm{ probability \: = \dfrac{no \: of \: fav \: events}{total \: no \: of \: events} }}$

$\implies \boxed { \boxed{ \rm{probability \: = \: \: \dfrac{1}{5} }}}$