MATHEMATICS – Class 10
PROBABILITY
1. In a cricket match probability of winning of India against Pakistan is 0.79. Then probability
of loosing the match will be
(A) 0.23 (B) 0.21 (C) 0.14 (D) 0.36
SOL : Probability of losing the match = 1 - 0.79 = 0.21
ANS : B
2. Probability that a non leap year should have 53 Mondays, will be
(A) (B) (C) (D)
SOL : In a non leap year there are 365 days. So there are 52 weeks and 1 days.
Probability of 53 mondays in a non loop year = 1/7.
ANS : C
3. A number is chosen randomly among the first 100 natural numbers. Then the probability
that the number chosen is multiple of 7, will be
(A) (B) (C) (D)
SOL : Probability of getting a multiple of 7 is
= 14/100 = 7/50.
ANS : A
4. A bag contains 10 red balls and some white balls. If the probability of drawing a white ball
is double that of a red ball, then number of white balls in the bag will be
(A) 10 (B) 15 (C) 20 (D) 25
SOL : Let the number of white balls in the bag be x.
Probability of getting a white ball = x / (10 + x)
Probability of getting a red ball = 10 / (10 + x)
now,
x = 20.
ANS : C
,5. Each outcome of a sample space related to any random experiment is known as
(A) compound event (B) elementary event (C) sure event (D) impossible event
ANS : B
6. The king, queen and jack of hearts are removed from a deck of 52 playing cards and then
well shuffled. One card is selected from the remaining cards. Then the probability of getting
a king is
(A) 1/49 (B) 2/49 (C) 3/49 (D) 1.
SOL : The king, queen and jack of hearts are removed from a deck of 52 playing cards.
Number of cards left in the deck = 52 – 3 = 49
Well shuffling of the cards ensures equally likely outcomes.
Out of 49 well shuffled cards, one card can be drawn in 49 ways.
total number of possible outcomes = 49.
There are 3 kings left in the remaining 49 playing cards as the king of the heart has been
removed.
Thus, the number of favourable ways of ‘getting a king’ = 3
Hence, P(a king) = 3/49
ANS : C
7. 12 defective pens are academically 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 taken out is a good one.
(A) 11/12 (B) 1/12 (C) 10/12 (D) 1/4.
SOL : Total number of pens = 132 + 12 = 144
Out of 144 pens, one pen can be taken out in 144 ways.
Total number of possible outcomes = 144
Number of good pens = 132.
Thus, the number of favourable ways of ‘getting a good pen’ = 132
Hence, .
ANS : A
8. 17 cards numbered 1, 2, 3, …. 16, 17 are put in a box and mixed thoroughly. One person
draws a card from the box. Then the probability that the odd number on the card is
(A) 8/17 (B) 9/17 (C) 6/17 (D) 5/17
SOL : Possible outcomes are 1, 2, 3, …. 16, 17
Total number of possible outcomes = 17
Odd numbers from 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17
Thus, the number of favourable ways of ‘getting an odd number on the card’ = 9
, Hence, P(odd number) = 9/17
ANS : B
9. Two coins are tossed simultaneously. Then the probability of getting at list one head is
(A) 1/4 (B) 1 (C) 3/4 (D) 0.
SOL : If two coins are tossed simultaneously then the possible outcomes are HH, HT, TH, TT.
Total number of possible outcomes = 4
Outcomes, favourable to the event ‘at least one head’ are HH, HT, TH
Thus, the number of favourable outcomes = 3
Hence, P(at least one head) = 3/4.
ANS : C
10. A die is thrown twice. Then the probability that 5 will come up at least once is
(A) 11/36 (B) 7/36 (C) 5/36 (D) 0.
SOL : When two dice are thrown, the possible outcomes are
(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)
Total number of possible outcomes = 36
Outcome favourable to the event ‘5 will come up at least once’ are
(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5)
Thus, the number of favourable ways of ‘5 coming up at least once’ = 11
Hence, P(5 will come up at least once) = 11/36
ANS : A
11. Ina single throw of two dice, then the probability of getting a doublet of odd numbers is
(A) 11/12 (B) 1/12 (C) 5/12 (D) 5/6.
SOL : When two dice are thrown, the possible outcomes are
(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)
PROBABILITY
1. In a cricket match probability of winning of India against Pakistan is 0.79. Then probability
of loosing the match will be
(A) 0.23 (B) 0.21 (C) 0.14 (D) 0.36
SOL : Probability of losing the match = 1 - 0.79 = 0.21
ANS : B
2. Probability that a non leap year should have 53 Mondays, will be
(A) (B) (C) (D)
SOL : In a non leap year there are 365 days. So there are 52 weeks and 1 days.
Probability of 53 mondays in a non loop year = 1/7.
ANS : C
3. A number is chosen randomly among the first 100 natural numbers. Then the probability
that the number chosen is multiple of 7, will be
(A) (B) (C) (D)
SOL : Probability of getting a multiple of 7 is
= 14/100 = 7/50.
ANS : A
4. A bag contains 10 red balls and some white balls. If the probability of drawing a white ball
is double that of a red ball, then number of white balls in the bag will be
(A) 10 (B) 15 (C) 20 (D) 25
SOL : Let the number of white balls in the bag be x.
Probability of getting a white ball = x / (10 + x)
Probability of getting a red ball = 10 / (10 + x)
now,
x = 20.
ANS : C
,5. Each outcome of a sample space related to any random experiment is known as
(A) compound event (B) elementary event (C) sure event (D) impossible event
ANS : B
6. The king, queen and jack of hearts are removed from a deck of 52 playing cards and then
well shuffled. One card is selected from the remaining cards. Then the probability of getting
a king is
(A) 1/49 (B) 2/49 (C) 3/49 (D) 1.
SOL : The king, queen and jack of hearts are removed from a deck of 52 playing cards.
Number of cards left in the deck = 52 – 3 = 49
Well shuffling of the cards ensures equally likely outcomes.
Out of 49 well shuffled cards, one card can be drawn in 49 ways.
total number of possible outcomes = 49.
There are 3 kings left in the remaining 49 playing cards as the king of the heart has been
removed.
Thus, the number of favourable ways of ‘getting a king’ = 3
Hence, P(a king) = 3/49
ANS : C
7. 12 defective pens are academically 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 taken out is a good one.
(A) 11/12 (B) 1/12 (C) 10/12 (D) 1/4.
SOL : Total number of pens = 132 + 12 = 144
Out of 144 pens, one pen can be taken out in 144 ways.
Total number of possible outcomes = 144
Number of good pens = 132.
Thus, the number of favourable ways of ‘getting a good pen’ = 132
Hence, .
ANS : A
8. 17 cards numbered 1, 2, 3, …. 16, 17 are put in a box and mixed thoroughly. One person
draws a card from the box. Then the probability that the odd number on the card is
(A) 8/17 (B) 9/17 (C) 6/17 (D) 5/17
SOL : Possible outcomes are 1, 2, 3, …. 16, 17
Total number of possible outcomes = 17
Odd numbers from 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17
Thus, the number of favourable ways of ‘getting an odd number on the card’ = 9
, Hence, P(odd number) = 9/17
ANS : B
9. Two coins are tossed simultaneously. Then the probability of getting at list one head is
(A) 1/4 (B) 1 (C) 3/4 (D) 0.
SOL : If two coins are tossed simultaneously then the possible outcomes are HH, HT, TH, TT.
Total number of possible outcomes = 4
Outcomes, favourable to the event ‘at least one head’ are HH, HT, TH
Thus, the number of favourable outcomes = 3
Hence, P(at least one head) = 3/4.
ANS : C
10. A die is thrown twice. Then the probability that 5 will come up at least once is
(A) 11/36 (B) 7/36 (C) 5/36 (D) 0.
SOL : When two dice are thrown, the possible outcomes are
(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)
Total number of possible outcomes = 36
Outcome favourable to the event ‘5 will come up at least once’ are
(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5)
Thus, the number of favourable ways of ‘5 coming up at least once’ = 11
Hence, P(5 will come up at least once) = 11/36
ANS : A
11. Ina single throw of two dice, then the probability of getting a doublet of odd numbers is
(A) 11/12 (B) 1/12 (C) 5/12 (D) 5/6.
SOL : When two dice are thrown, the possible outcomes are
(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)
