CBSE Class 10 Maths Chapter-15 Probability – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 15 – Probability prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-15 Probability Important Questions
CBSE Class 10 Maths Important Questions Chapter 15 – Probability
4 Mark Questions
1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Ans. Total number of favourable outcomes = 52
(i) There are two suits of red cards, i.e., diamond and heart. Each suit contains one king.
Favourable outcomes = 1
Hence, P (a king of red colour) = 
(ii) There are 12 face cards in a pack.
Favourable outcomes = 12
Hence, P (a face card) = 
(iii) There are two suits of red cards, i.e., diamond and heart. Each suit contains 3 face cards.
Favourable outcomes = 2 x 3 = 6
Hence, P (a red face card) = 
(iv) There are only one jack of heart.
Favourable outcome = 1
Hence, P (the jack of hearts) = 
(v) There are 13 cards of spade.
Favourable outcomes = 13
Hence, P (a spade) = 
(vi) There is only one queen of diamonds.
Favourable outcome = 1
Hence, P (the queen of diamonds) =
2. 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?
Ans. (i) The outcomes associated with the experiment in which a dice is thrown is twice:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Therefore, Total number of favourable outcomes = 36
Now consider the following events:
A = first throe shows 5 and B = second throw shows 5
Therefore, the number of favourable outcomes = 6 in each case.
P(A) = 
and P(B) =
P
and P
Required probability = 
(ii) Let S be the sample space associated with the random experiment of throwing a die twice. Then, n(S) = 36
A
B = first and second throw shoe 5, i.e. getting 5 in each throw.
We have, A = (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
And B = (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)
P(A) = , P(B) = and P(AB) = 
Required probability = Probability that at least one of the two throws shows 5
= P (A
B) = P(A) + P(B) – P(AB)
= 
3. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Ans. Total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Exta are:
(T, T) (T, W) (T, TH) (T, F) (T, S)
(W, T) (W, W) (W, TH) (W, F) (W, S)
(TH, T) (TH, W) (TH, TH) (TH, F) (TH, S)
(F, T) (F, W) (F, TH) (F, F) (F, S)
(S, T) (S, W) (S, TH) (S, F) (S< S)
Total number of favourable outcomes = 25
(i) The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S).
Number of favourable outcomes = 5
Hence required probability = 
(ii) The favourable outcomes of visiting on consecutive days are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).
Number of favourable outcomes = 8
Hence required probability = 
(iii) Number of favourable outcomes of visiting on different days are 25 – 5 = 20
Number of favourable outcomes = 20
Hence required probability = 
4. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is
(i) a card of spades of an ace
(ii) a red king
(iii) neither a king nor a queen
(iv) either a king or a queen
(v) a face card
(vi) cards which is neither king nor a red card.
Ans. Total possible outcomes = 52
(i) No. of spades = 13
No. of ace = 4
1 card is common [ace of spade]
Favourable outcomes = 13+4 – 1=16
Required probability = 
(ii) No. of red kings = 2
Favourable outcomes = 2
Required probability = 
(iii) No. of king and queen = 4+4=8
Favourable outcomes = 52 – 8=44
Required probability = 
(iv) No. of king and queen =4+4=8
Required probability = 
(v) No. of face cards = 4+4+4=12 [Jack, queen and king are face card]
Required probability = 
(vi) No. of cards which are neither red card nor king = 52 – (26+4 – 2
=52 – 28=24
Required probability = 
5. Cards marked with numbers 1,2,3,…25 are placed in a box and mixed thoroughly and one card is drawn at random from the box, what is the probability that the number on the card is
(i) a prime number?
(ii) a multiple of 3 or 5?
(iii) an odd number?
(iv) neither divisible by 5 nor by 10?
(v) perfect square?
(vi) a two-digit number?
Ans. Total no. of possible outcomes = 25
(i) favourable cases are 2,3,5,7,11,13,17,19,23 which are 9 in number
Required probability = 
(ii) Multiple of 3 or 5
Favourable cases are 3,5,6,9,10,12,15,18,20,21,24,25, which are 12 in number
Required probability = 
(iii) Favourable cases are 1,3,5,7,9,11,13,15,17,19,21,23,25, which are 13 in number
Required probability = 
(iv) Favourable cases are 1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24, which are 20 in number
Required probability = 
(v) Perfect square numbers are 1,4,9,16,25
Favourable cases are = 5
Required probability = 
(vi) Two-digit numbers are 10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25 =16
Required probability = 
6. From a pack of 52 playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is (i) a black queen, (ii) a red card, (iii) a black jack,(iv) a honorable card.
Ans. Total number of outcomes = 52
Cards removed = 2+2+2+2 = 8[2 jack, 2 queen, 2 king and 2 aces of red colour]
Remaining number of cards = 52 – 8=44
Total number of outcomes = 44
(i) Favourable outcomes = 2 [There are 2 black queen]
Required probability = 
(ii) Favourable outcomes = number of red cards left = 26 – 8=18
Probability for a red card = 
(iii) Favourable outcomes = Number of black jacks = 2
Required probability =
(iv) Number of picture cards left =2+2+2 = 6 [jack, queen, King are picture cards]
Required probability = 
(v) Honorable cards [ace, jack, queen and king]
No. of honorable cards left = 2+2+2+2=8
Required probability = 