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1. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen take out is a good one.
Number of defective pens = 12
Number of non - defective pens = 132
Total number of pens = 12 + 132 = 144
Let 'A' be the favorable outcomes of getting the pen which is a good one. Then,
n (A) = 132
Therefore, P (A) = n (A)/ n (S) = 132/ 144 = 11/12.
2. Suppose you drop a die at random on the rectangular region shown in the picture. What is the probability that it will tend inside the circle with diameter 1 m ?
Area of rectangular region = 3 x 2 = 6 m2
Diameter of the circle = 1 m
∴ Radius = 1/2 m
So, Area = πr2 = π ( 1/2 ) 2 = π/ 4 m2
Now, Probability that the die will land inside the circle = 
3. A die is thrown twice. What is the probability that
(i) 5 will not come up either time ?
(ii) 5 will come up at least once ?
When a die is thrown twice, then possible outcomes are
i.e. n (S) = 36.
( i ) Let A be the favorable outcomes that 5 will not come up either time.
(ii) Let B be the favorable outcomes that 5 will come at least once. Then
4. A box contains cards marked with numbers 5 to 20. A card is drawn from the bag at random. Find the probability of getting a number which is a perfect square.
Total number of possible outcomes are 16.
ie, n (s) = 16
Let A be the favorable outcomes of getting a number which is a perfect square. Then A = (9, 16)
ie, n (A) = 2
Therefore,
⇒ P (A) = n ( A ) / n (S ) = 2/16 = 1/8
5. A die is thrown once. Find the probability of getting a number less than 3.
When a die is thrown once then possible outcomes are 6.
ie. n (S) = 6
Let A be the favorable outcomes of getting a number less than 3 then
A = (1, 2)
ie. n (A) = 2
⇒ P (A) = n ( A ) / n (S ) = 2/6 = 1/3
6. Two friends were born in the year 2000. What is the probability that they have the same birthday ?
2000 is a leap year. So, their birthday can be any day of 366 days in the year.
We assume that these 366 outcomes are equally likely. Because both the friends have the same birthday.
∴ favorable outcomes for their birthday = 1
Hence, P(both have the same birthday) = 1/ 366.
7. Which of the following can not be the probability of an event ?
(A) 2/3 (B) - 1.5
(C) 15 % (D) 0.7
As we know probability of event can not be less than 0 and greater than 1.
i,e. 0 ≤ P ≤ 1
∴ (B) - 1. 5 is not possible.
8. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. Find the probability of getting a black ball.
No. of red balls = 4
No. of black balls = 6
Total no. of balls 4 + 6 = 10
Let A be the favorable out comes of getting a black ball, then, n ( A ) = 6
Therefore,
P (A ) = n (A ) / n (S ) = 6/10 = 3/5.
9. From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
When a Card is drawn of random from a well shuffled pack of cards, then possible outcomes are 52.
i.e. n (S) = 52
Let A be the favorable outcomes of getting a black queen, then n (A) = 2
Therefore, P (A) = n (A) / n (S) = 2/52 = 1/26 .
10. 12 Cards bearing numbers 3 to 20 are placed in a bag and mixed. A card is taken out from the bag at random. What is the probability that the number on the card taken out is an even number.
Total no. of possible outcomes are 18.
i.e, n (S) = 18
Let A be the favorable outcomes of getting the number on the card which is an even number, then
A = (4, 6, 8, 10, 12, 14, 16, 18, 20)
i.e., n (A) = 9
Therefore, P(A ) = n (A) / n (S) = 9/18 = 1/2 .
11. A die is thrown once. Find the probability of getting
(i) an even prime number
(ii) a multiple of 3
If we throw a die, the possible outcomes are
S = {1, 2, 3, 4, 5, 6} i.e. n (S) = 6
(i) Let A be the favorable outcomes of getting an even prime number, then
A = (2) i.e. n (A) = 1
Therefore, P (A) = n ( A )/ n ( s ) = 1/6
(ii) Let B be the favorable outcomes of getting a multiple of three, then
B = (3, 6) i.e., n (B) = 2
Therefore, P (B) = n (B)/n (S ) = 2/6 = 1/3
12. A card is drawn from a pack of 52 playing cards. What is the probability that it is
(i) an ace
(ii) a face card
(iii) any card numbered from 2 to 10.
Total number of all possible outcomes = 52
i. e, n (S) = 52
(i) Let E be the favorable outcomes of getting an ace, then n (E) = 4
Therefore, P (E) = n (E)/ n (S) = 4/52 = 1/13.
(ii) Let F be the favorable outcomes of getting a face card, then
n (E) = 12
Therefore,
P (E) = n ( F )/ n ( S ) = 12/52 = 3/13.
(iii) Let G be the favorable outcomes of getting any card numbered from 2 to 10, then n (G) = (36).
Therefore, P (G) = n (G)/ n (S) = 36/52 = 9/13.
13. A card is drawn at random from a well - shuffled deck of playing cards. Find the probability of drawing a (i) face card (ii) Card which is neither a king nor a red card.
If one card is drawn from a well - shuffled deck of playing card, then possible outcomes are 52.
i. e., n (S) = 52
I. Let A be the favorable outcomes of getting a face card, then
n (A) = 12
Therefore, P ( A ) = n ( A )/ n ( S ) = 12/ 52 = 3/ 13.
II. Let B be the favorable outcomes of getting card which is neither a king nor a red card, then n (B) = 24.
Therefore, P (B) = n (B )/ n ( s ) = 24/52 = 6/ 13.
14. The king, Queen and jack of clubs are removed from a deck of 52 cards and then well shuffled. One card is selected from the remaining cards. Find the probability of getting (i) a heart (ii) a king (iii) a club (iv) the '10' of heart.
If the king, queen and jack of clubs are removed from a deck of 52 cards, then possible outcomes become 49.
n (S) = 49 [ 52 - 3 = 49 ]
(i) Let E be the favorable outcomes of getting a heart then, n (E) = 13
Therefore, P (E) = n (E )/n ( S ) = 13/49
(ii) Let F be the favorable outcomes of getting a king then, n (E) = 3
Therefore, P (E) = n (E )/n ( S ) = 3/49
(iii) Let G be the favorable outcomes of getting a club then, n (G) = 10
Therefore, P(G) = n (G )/n ( S ) = 10/49
(iv) Let H be the favorable outcomes of getting the 10 of hearts, then n (H) = 1
Therefore, P (H) = n (H )/n ( S ) = 1/49.
15. A king, queen and jack of diamonds are removed from a pack of 52 cards and then the pack is well shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) diamonds, (ii) a jack.
When a king, queen and jack of diamonds are removed from a pack of 52 cards, then possible outcomes are 52 - 3 = 49
i.e., n (S) = 49
i. Let A be the favorable outcomes of getting a card of diamonds. Then n (A) = 10
Therefore, P (A) = n (A)/ n (S) = 10/ 49.
ii. Let B be the favorable outcomes of getting a jack. Then n (B) = 3.
Therefore, P (B) = n (B)/ n (S) = 3/ 49.
16. All the three cards of spades are removed from a well - shuffled pack of 52 cards. A card is drawn at random from the remaining pack. Find the probability of getting
(a) a black face card (b) a black card.
If all the three face cards of spades are removed, then possible outcomes are 49.
(a) Let A be the favorable outcomes of getting a black face card, then, n (A) = 3.
Therefore,P (A) = n ( A )/ n ( S ) = 3/49 .
(b) Let C be the favorable outcomes of getting a black card, then, n (C) = 23
Therefore, P (C) = n (C) / n (S) = 23 /49.
17. Red kings, queens and jacks are removed from a deck of 52 playing cards and then well shuffled. A card is drawn from the remaining cards. Find the probability of getting
(i) a king, (ii) a red card, (iii) a spade.
If Red kings, queens and jacks are removed from a deck of 52 cards, then, n (S) = 52
(i) Let 'A' be the favorable outcomes of getting a king, then n (A) = 2
Therefore, P (A) = n (A) / n (S) = 2/42 = 1/ 21.
(ii) Let 'B' be the favorable outcomes of getting ' red card' then, n (B) = 20
Therefore, P (B) = n (B )/ n ( S ) = 20/42 = 10/ 21.
(iii) Let 'C ' be the favorable outcomes of getting a spade then, n (C) = 11
Therefore, P (C) =n (C)/ n (S) = 11/ 42.
18. A bag contains 4 red, 5 black and 3 yellow balls. A ball is taken out of the bag at random. Find the probability that the ball taken out is of (i) yellow colour, (ii) not of red colour.
No. of red balls = 4
No. black balls = 4
No of yellow balls = 3
Total no. of balls = (4 + 5 + 3) = 12
n (S) = 12
I. Let 'A ' be the favorable outcomes of getting the ball taken out is of ' yellow colour'. Then, n (A) = 3
Therefore, P (A) = n ( A )/ n ( S ) = 3/12 = 1/4.
Ii Let 'B' be the favorable outcomes of getting the ball taken out is of red color. Then, n (B) = 8
Therefore, P (B) = n (B)/ n (S) = 8/12 = 2/3.
19. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is four times that of red ball. Find the number of blue balls in the bag.
Number of red balls in the bag = 5
let number of blue balls in the bag = x
Now, total number of balls in the bag = x + 5
i.e., n (S) = x + 5
(i) Let 'A' be the favorable outcomes of getting blue balls, then, n (A) = x
Therefore, P(A ) = n (A)/ n (S) = x / (x + 5)
(ii) Let ' B ' be the favorable outcomes of getting red balls, then, n (B) = 5
Therefore, P(B) = n (B)/ n (S) = 5 / (x + 5)
According to question:
P (A) = 4 P (B)
⇒ x / x + 5 = 4 ( 5/ (x + 5) )
⇒ x = 20
Hence, the number of blue balls in the bag = 20
20. A bag contains 5 red, 8 green and 7 white balls. On ball is drawn at random from the bag, find the probability of getting
(i) a white ball or a green ball.
(ii) neither a green ball not a red ball.
Total number of balls in the bag = 5 + 8 + 7 = 20.
i.e. n (S) = 20.
(i) Let A be the favorable outcomes of getting a white ball or a green ball. Then, n (A) = 15
Therefore, P (getting a white ball or a green ball) = n ( A )/ n ( S ) = 15/ 20 = 3/4
(ii) Let B be the favorable outcomes of getting neither of a green ball nor a red ball. Then, n (B) = 7
Therefore, P (getting neither a green ball nor a red ball) = n ( B ) / n ( s ) = 7/ 20.
21. 12 cards, numbered 1, 2, 3 ........,12 are put in a box and mixed. A card is drawn at random from the box. Find the probability that the card drawn bears.
( i ) an even number
( ii ) a number divisible by 2 or 3.
Total number of cards = 12
i. e., n (S) = 12
(i) Let A be the favorable outcomes of getting an even number, then A = { 2, 4, 6, 8, 10, 12 }
⇒ n (A) = 6
Therefore, P (A) = n ( A ) /n ( s ) = 6/12 = 1/2
(ii) Let B the favorable outcomes of getting a number divisible by 2 or 3 , then, n (B) = 8
Therefore, P (B) = n ( B )/n ( s ) = 8/12 = 2/3.
22. A bag contains tickets, number 11, 12, 13 ----------- 30. A ticket is taken out form the bag at random. Find the probability that the number on the drawn ticket is ( i ) a multiple of 7, (ii) greater than 15 and a multiple of 5.
Total Number of tickets = 20
i.e., n (S) = 20
(i) Let 'A' be the favorable outcomes of getting multiple of 7, then, A = (14, 21, 28)
i . e., n (A) = 3
Therefore, P (A) = n (A)/n (S) = 3/20 .
(ii) Let 'B' be the favorable outcomes of getting the number greater than 15 and multiple of 5, then, B = (20, 25, 30)
i.e., n (B) = 3
Therefore, P (B) = n ( B )/ n ( S ) = 3/20
23. Fill in the blanks:
1. Probability of an event E + Probability of the event 'not E' = _________.
2. A die is thrown once, the probability of getting a prime number is ________.
3. If the probability of an event of a random experiment is p(E) = 0, then the event is called ________.
4. The probability of an event that is certain to happen is ________ such an event is called ________.
1. 1
2. 1/2
3. impossible event
4. 1, sure or certain event
24. Match the proposed probability under column I with the appropriate written description under column II.
Column I Column II
Probability Written Description
(A) 0.95 (p) An incorrect assignment
(B) 0.02 (q) Very little chance of happening
(C) - 0.3 (r) As much chance of happening as not
(D) 0.5 (s) Very likely to happen
(A) → (s) ; (B) → (q) ; (C) → (p) ; (D) → (r)
25. A jar contains only green, white and yellow marble. The probability of selecting a green marble and white marble randomly from a jar is 1/4 and 1/3 respectively. If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?
Let the number of green marbles = x.
Let the number of white marbles = y.
∴ Total number of marbles = x + y = 10.
Now, P (selecting a green marble) = x / x + y + 10.
But it is given that P (green marble) = 1 / 4
∴ x / x + y + 10 = 1 / 4
4x = x + y + 10
3x - y = 10 .......... (1)
Similarly, P(selecting a white marble) = y / x + y + 10
∴ y / x + y + 10 = 1 / 3
2y - x = 10 ............(2)
On Solving the equation (1) and (2) we get, y = 8 and x = 6.
∴ Total number of marbles = x + y + 10 = 24.
26. An unbiased die is rolled twice. Find the probability of getting (i) the sum of two numbers as a prime (ii) the sum of two numbers equal to 9.
(i) The sum of the two numbers lies between 2 and 12. So the primes are 2, 3, 5, 7, 11.
No. of ways of getting 2 = (1, 1) = 1
No. of ways getting 3 = (1, 2), (2, 1) = 2
No. of ways getting 5 = (1, 4), (4, 1), (2, 3), (3, 2) = 4
No. of getting 7 = (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6
No. of ways getting 11 = (5, 6), (6, 5) = 2
No. of favorable ways = 6 x 6 = 36.
Probability of the sum as a prime = 15 / 36 = 5 / 12
(ii) No. of ways of getting a sum of 9 = (3, 6), (6, 3), (4, 5), (5, 4) = 4
probability of getting a sum of 9 = 4 / 36 = 1 / 9.
