## Filters

Clear All

H Harsh Kankaria
Total number of pens = 144 Total number of defective pens = 20  Number of good pens = 144-20 = 124 She will buy if the pen is good. Therefore, the probability that she buys = probability that the pen is good =

H Harsh Kankaria
Here, Total outcome is the area of the rectangle and favourable outcome is area of the circle. Area of the rectangle =  Area of the circle =

H Harsh Kankaria
The six faces of the die contains : {A,B,C,D,E,A} Total number of letters = 6 (i) Since there is only one D, number of favourable outcomes = 1 Therefore, the probability of getting D is

H Harsh Kankaria
The six faces of the die contains : {A,B,C,D,E,A} Total number of letters = 6 (i) Since there are two A's, number of favourable outcomes = 2 Therefore, the probability of getting A is

H Harsh Kankaria
Total number of discs = 90 Numbers between 1 and 90 that are divisible by 5 are {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90} Therefore, total number of discs having numbers that are divisible by 5 = 18.

H Harsh Kankaria
Total number of discs = 90 Perfect square numbers between 1 and 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81} Therefore, total number of discs having perfect squares = 9.

H Harsh Kankaria
Total number of discs = 90 Number of discs having a two-digit number between 1 and 90 = 81

H Harsh Kankaria
Total number of bulbs =  20 Hence, total possible outcomes = 20 Number of defective bulbs = 4 Hence, the number of favourable outcomes = 4

H Harsh Kankaria
Total number of bulbs =  20 Hence, total possible outcomes = 20 Number of defective bulbs = 4 Hence,number of favourable outcomes = 4

H Harsh Kankaria
Total number of pens = 132(good) + 12(defective) Hence, total possible outcomes = 144 Number of good pens = number of favourable outcomes = 132

H Harsh Kankaria
When the queen is kept aside, there are only 4 cards left Hence, the total possible outcomes = 4 (2b) Since there is no queen left. Hence, favourable outcome = 0 Therefore, the probability of getting a queen is 0. Thus, it is an impossible event.

H Harsh Kankaria
When the queen is kept aside, there are only 4 cards left Hence, the total possible outcomes = 4 (2a) There is only one ace. Hence, favourable outcome = 1 Therefore, the probability of getting an ace is 0.25

H Harsh Kankaria
Total number of cards = 5 Hence, the total possible outcomes = 5 (1) There is only one queen. Hence, favourable outcome = 1

H Harsh Kankaria
Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (6) Let E be the event of getting the queen of diamonds Hence, the number of favourable outcomes = 1 Therefore, the probability of getting the queen of diamonds is

H Harsh Kankaria
Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (5) Let E be the event of getting a spade. There are 13 cards in each suit. {2,3,4,5,6,7,8,9,10,J,Q,K,A} Hence, number of favourable outcomes = 13 Therefore, the probability of getting a spade is

H Harsh Kankaria
Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (4) Let E be the event of getting the jack of hearts Hence, the number of favourable outcomes = 1 Therefore, the probability of getting the jack of hearts is

H Harsh Kankaria
Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (3) Let E be the event of getting a red face card. Face cards: (J, Q, K) of hearts and diamonds Hence, number of favourable outcomes = 3x2 = 6 Therefore, the probability of getting a red face card is

H Harsh Kankaria
Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (2) Let E be the event of getting a face card. Face cards: (J, Q, K) of each four suits Hence, number of favourable outcomes = 12 Therefore, the probability of getting a face card is