Probability Class 10 MCQ with Solution
1. A bag contains 8 black balls and some white balls. If the probability of drawing a white ball is twice that of a black ball, then find the number of white balls in the bag.
(A) 8
(B) 10
(C) 12
(D) 16
Ans. (D)
Sol. Let number of white balls in the bag = x
And Number of black balls in the bag = 8 (‡ given)
Total number of balls in the bag = (8 + x)
i.e. Total possible outcomes = (8 + x) [say n(s)]
Let E be the event of getting white ball and F be the event of getting black ball.
n(E) = no. of favourable case of white = x
n(F) = no. of favourable case of black ball = 8
According to question
Probability of drawing a white ball
= 2 × Probability of drawing black ball
P(E) = 2 P(F)
\frac{{n({\bf{E}})}}{{n(s)}} =2 \times \frac{{n({\bf{F}})}}{{n(s)}}
\frac{x}{{(8 + x)}} =\frac{{2 \times 8}}{{8 + x}}
x =16 \times \frac{{(8 + x)}}{{(8 + x)}}
x = 16
i.e. the number of white balls in the bag = 16.
2. What is the probability that a leap year has 53 sundays and 53 mondays both ?
(A) \frac{1}{7}
(B) \frac{2}{7}
(C) 0
(D) \frac{1}{2}
Ans. (A)
Sol. Total days in a leap year = 366 days
Number of weaks = 366 ¸ 7
= 52 weaks + 2 days
The 52 weaks means every day is 52 times in a year
now left 2 days will be (Sun-Mon), (Mon-Tue), (Tue-Wed), (Wed-Thr), (Thr-Fri), (Fri-Sat), (Sat-Sun)
Total possible out comes = 7 [say n(s)]
Favourable case 53 sundays and 53 mondays both i.e. n(E) = 1
P(E) =\frac{{n({\bf{E}})}}{{n(s)}} = \frac{1}{7}
3. In a single throw with two dice the odds against drawing 7 is :
(A) \frac{1}{6}
(B) \frac{1}{{12}}
(C) 5 : 1
(D) 1 : 5
Ans. (C)
Sol. Odds against means =\frac{{{\bf{no}}{\bf{.}}\,\,{\bf{of}}\,\,{\bf{unfavourable}}\,\,{\bf{case}}}}{{{\bf{no}}{\bf{.}}\,\,{\bf{of}}\,\,{\bf{favourable}}\,\,{\bf{case}}}}
When a single throw with two dice
i.e. Total possible outcomes = n(s) = 36
and drawing 7
favourable case n(E) = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
i.e. no. of favourable case = 6
no. of unfavourable case = Total case – no. of favourable case
= 36 – 6 = 30
odd’s against drawing 7 =\frac{{30}}{6} = \frac{5}{1} = 5:1
4. Three dice are thrown simultaneously. The probability of getting a sum of 15 is :
(A) \frac{1}{{72}}
(B) \frac{5}{{36}}
(C) \frac{5}{{72}}
(D) None of these
Ans. (D)
Sol. Three dice are thrown simultaneously
The total possible outcomes = 6 × 6 × 6 = 216
Probability of getting a sum of 15 is then favourable cases
= {(6, 6, 3), (6, 5, 4), (6, 4, 5), (6, 3, 6), (3, 6, 6), (5, 4, 6), (4, 6, 5), (5, 6, 4), (5, 5, 5), (4, 5, 6)}
Number of favourable cases = 10
Probability of getting sum 15 is = \frac{{10}}{{216}} i.e. ans. is (D)
5. Three dice are thrown. The probability of getting a sum which is a perfect square is :
(A) \frac{2}{5}
(B) \frac{9}{{20}}
(C) \frac{1}{4}
(D) None of these
Ans. (D)
Sol. Three dice are thrown
Total possible outcomes = 216.
The number of favourable case i.e. getting a sum which is perfect square are
Number of favourable case = 13
Probability getting a sum which is a perfect square
=\frac{{{\bf{favourable}}\,\,{\bf{case}}}}{{{\bf{total}}\,\,{\bf{possible}}\,\,{\bf{out}}\,\,{\bf{comes}}}}
= \frac{{13}}{{216}} i.e. answer (D)
6. A dice is thrown once. The probability of getting a number less than 6 is:
(A) zero
(B)\frac{1}{2}
(C)\frac{5}{6}
(D) none of these
Ans. (C)
Sol. Required probability =\frac{5}{6} (favourable cases are getting 1, 2, 3, 4, 5)
7. A dice is thrown once. The probability of getting a number greater than 2 is:
(A)\frac{1}{2}
(B)\frac{1}{3}
(C)\frac{2}{3}
(D)\frac{3}{5}
Ans. (C)
Sol. Required probability =\frac{2}{3} (favourable cases are getting 3, 4, 5, 6)
8. What is the probability that a number selected at random from the number 1, 2, 2,3, 3, 3, 3, 4, 4, 4, 4 will be their average
(A) 4/11
(B) 3/11
(C) 5/11
(D) 7/11
Ans. (A)
Sol. Average of the given number =\frac{{1 + 2 + 2 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4}}{{11}} = 3
No. of favourable case = 4
P(a number average of the given number) =\frac{4}{{11}}
9. The probability of selecting a green marble at random from a Jar that contains only green, white and yellow marbles is 1/4. The probability of selecting a white marble at random from the same Jar is 1/3. If this jar contains 10 yellow marble. What is the total number of marble in the jar.
(A) 21
(B) 16
(C) 24
(D) 18
Ans. (C)
Sol. P(E1) + P(E2) + P(E3) = 1 [‡ jar contain only green, white and yellow marbles] \frac{1}{4} + \frac{1}{3} + \frac{{10}}{x} = 1 \frac{{10}}{x} = \frac{5}{{12}} \Rightarrow \,\,x = 24
10. There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written, one card is taken out of the bag at random, find the probability that the number on the selected card is not divisible by 3.
(A) 1/3
(B) 2/3
(C) 1/5
(D) 2/5
Ans. (B)
Sol. The total number of cards = 30 [1 to 30]
i.e. total possible outcomes = 30
The no. which divisible by 3 = {3, 6, 9, . . . ., 30} = i.e. 10
Hence the number which are not divisible by 3 = 30 – 10 = 20
P (not divisible by 3) =\frac{{20}}{{30}} = \frac{2}{3}
i.e. Answer (B)
11. In a simultaneous throw of a pair of dice. The probability of getting a doublet is :
(A) \frac{1}{6}
(B) \frac{5}{6}
(C) \frac{2}{3}
(D) \frac{1}{3}
Ans. (A)
Sol. Total possible outcomes = 36
Favourable Outcomes = 6{(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
P(E) =\frac{6}{{36}} = \frac{1}{6}
12. A card is drawn from a well shuffled deck of 52 cards then probability of getting either a red card or a queen is :
(A) \frac{7}{{13}}
(B) \frac{8}{{13}}
(C) \frac{9}{{13}}
(D) \frac{5}{{13}}
Ans. (A)
Sol. Total favourable cards = 26 + 2 = 28
P(E) =\frac{{28}}{{52}} = \frac{7}{{13}}
13. A card is drawn from a well shuffled deck of 52 cards then probability of gettting a face card is :
(A) \frac{3}{{13}}
(B) \frac{4}{{13}}
(C) \frac{5}{{13}}
(D) \frac{1}{{13}}
Ans. (A)
Sol. Total face cards = 12
P(F) =\frac{{12}}{{52}} = \frac{3}{{13}}
14. Two coins are tossed simultaneously then the probability of getting at least one tail is :
(A) 1
(B) \frac{1}{2}
(C) \frac{1}{4}
(D) \frac{3}{4}
Ans. (D)
Sol. Total possible outcomes = 4 (H, H) (H, T) (T, H) (T, T)
At least one tail = 3
P(T) = \frac{3}{4}
15. A card is drawn from a well shuffled deck of 52 cards then probability of getting an honour card is :
(A) \frac{2}{{13}}
(B) \frac{3}{{13}}
(C) \frac{4}{{13}}
(D) \frac{5}{{13}}
Ans. (C)
Sol. Total Honour cards = 16
P(H) =\frac{{16}}{{52}} = \frac{4}{{13}}
16. The probability that an ordinary year selected at random will contain 53 Sundays is :
(A) \frac{1}{7}
(B) \frac{2}{7}
(C) \frac{3}{7}
(D) None of these
Ans. (A)
Sol. In an ordinary year = 365 days
⇒ 52 Weeks + 1 day
Remaining one day can be Sun, Mon, Tues, Wed, Thurs, Fri, Satur.
P(53 Sun) = \frac{1}{7}
17. A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random,the probability that it is not green is:
(A) \frac{1}{6}
(B) \frac{{13}}{{24}}
(C) \frac{{11}}{{24}}
(D) \frac{5}{6}
Ans. (D)
Sol. P(\bar R) =\frac{{20}}{{24}} = \frac{5}{6}
18. A die is thrown once.What is the probability of getting a prime number ?
(A)\frac{1}{6}
(B)\frac{2}{3}
(C)\frac{1}{2}
(D) None of these
Ans. (C)
Sol. P(E) equals to\frac{3}{6}. (favorable cases are 2, 3, 5)
19. A die is thrown once. What is the probability of getting a number greater than 4 :
(A)\frac{1}{2}
(B) \frac{1}{3}
(C) \frac{2}{3}
(D) \frac{3}{5}
Ans. (B)
Sol. P(E) equals to \frac{1}{3}. (favorable cases are 5, 6)
20. A die is thrown once, what is the probability of getting a number less than 7 :
(A) zero
(B) \frac{1}{2}
(C) \frac{5}{6}
(D) None of these
Ans. (D)
Sol. The probability of getting a number less than 7 equals to 1.
21. A letter of the English alphabet is chosen at random. Find the probability that the letter chosen precedes P
(A) 15/26
(B) 11/26
(C) 4/13
(D) None of these
Ans. (A)
Sol. Total letter preced is ‘P’ are 15.
So, P(E) =\frac{{15}}{{26}}
22. From a set of 40 cards numbered 1,2, 3 …… 39, 40, one card is drawn at random. Find the probability that the drawn card bears a number which is divisible by both 3 and 5.
(A) 3/20
(B) 1/20
(C) 5/20
(D) 7/10
Ans. (B)
Sol. Favourable cases = 15, 30
P(E) =\frac{2}{{40}} = \frac{1}{{20}}
23. A bag contains some red balls and some white balls. If the probability of drawing a white ball is \frac{3}{5} then find the ratio of red to white balls.
(A) 1 : 2
(B) 2 : 3
(C) 2 : 5
(D) 5 : 6
Ans. (B)
Sol. Let no. of white balls = 3x
So total balls = 5x, so Red Balls = 2x
So, Ratio =\frac{{2x}}{{3x}} = \frac{2}{3}
24. A lady has 10 one-hundred notes in her purse. She takes out one note from her purse without looking into it. What is the probability that it is a fifty rupee note?
(A) 1/2
(B) 2/3
(C) 0
(D) 1
Ans. (C)
Sol. There were no fifity rupee notes. So probability equals to zero.
25. A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is \frac{2}{3}. Find the number of blue marbles in the jar.
(A) 7
(B) 9
(C) 8
(D) 5
Ans. (C)
Sol. Let no. of blue marbles = x
So, Green marbles = 24 – x
So, \frac{2}{3} =\frac{{24-x}}{{24}}
⇒ x = 8.
26. In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half minute after starting?
(A) 1/3
(B) 2/3
(C) 4/5
(D) 1/4
Ans. (D)
Sol. Total outcomes = 4 (4 slots in 2 mins.)
So P(E) =\frac{1}{4}
27. In a throw of a dice, the probability of getting an odd number is :
(A) 2
(B) \frac{1}{2}
(C) \frac{3}{2}
(D) 6
Ans. (B)
Sol. Total odd numbers = 3
So, P(E) =\frac{3}{6} = \frac{1}{2}
28. The probability that a vowel selected at random in English language is an “i” is :
(A) \frac{1}{5}
(B) \frac{1}{{26}}
(C) \frac{1}{6}
(D) None of these
Ans. (A)
Sol. Total vowels = 5(A, E, I, O, U)
So, P(E) =\frac{1}{5}
29. From a normal pack of cards, a card is drawn at random. The probability of getting a Queen or a King is
(A) \frac{2}{{52}}
(B) \frac{1}{{52}}
(C) \frac{2}{{13}}
(D) None of these
Ans. (C)
Sol. The probability of getting a king or Queen is.
P(E) =\frac{8}{{52}} = \frac{2}{{13}} (Total 4 Kings and 4 Queen)
30. When two dice are thrown, the probability of getting equal numbers is
(A) \frac{1}{{36}}
(B) \frac{1}{6}
(C) 1
(D) None of these
Ans. (B)
Sol. P(E) =\frac{6}{{36}} = \frac{1}{6}
fevourable cases = (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)
31. The probability of guessing the correct answer to a certain test questions is\frac{P}{5}. If the probability of not guessing the correct answer to this questions is\frac{3}{5}, then P =
(A) 2
(B) 3
(C) 4
(D) 6
Ans. (A)
Sol. \frac{P}{5} + \frac{3}{5} = 1
⇒ \frac{P}{5}=\frac{2}{5} ⇒ P = 2
32. A number is selected at random from the number 3,5,5,7,7,7,9,9, 9,9. The probability that the selected number is their average is
(A)\frac{1}{{10}}
(B)\frac{3}{{10}}
(C)\frac{7}{{10}}
(D)\frac{9}{{10}}
Ans. (B)
Sol. Average =\frac{{3 + 5 + 5 + 7 + 7 + 7 + 9 + 9 + 9 + 9}}{{10}} = 7
So P(E) =\frac{3}{{10}} [there are three 7’s out of ten numbers)
33. Amit throws three dice in a special game of Ludo. If it is known that he needs 15 or higher in this throw to win then find the chance of his winning the game.
(A)\frac{5}{{54}}
(B)\frac{{17}}{{216}}
(C)\frac{{13}}{{216}}
(D)\frac{{15}}{{216}}
Ans. (A)
Sol. P(E) =\frac{{20}}{{216}} = \frac{5}{{54}} [There are ‘20’ favourable chances]
34. If one card is missing from a pack of cards, what is the probability that it is a red ace :
(A)\frac{1}{{26}}
(B) \frac{1}{{32}}
(C) \frac{1}{2}
(D) \frac{1}{{52}}
Ans. (A)
Sol. P(E) =\frac{2}{{52}} = \frac{1}{{26}} [two red colour ace.]
35. A coin is tossed three times. The chance that head and tail show alternatively is :
(A)\frac{1}{8}
(B) \frac{3}{8}
(C)\frac{1}{4}
(D) \frac{3}{4}
Ans. (C)
Sol. P(E) =\frac{2}{8} = \frac{1}{4} [favourable cases are two, (HTH & THT)]
36. A coin is tossed two times. The probability of getting exactly two heads :
(A )\frac{1}{8}
(B) \frac{3}{8}
(C)\frac{1}{4}
(D) \frac{3}{4}
Ans. (C)
Sol. P(E) =\frac{1}{4} (favourable case is HH)
37. If one card is missing from a pack of cards, what is the probability that it is a black king :
(A)\frac{1}{{26}}
(B) \frac{1}{{32}}
(C) \frac{1}{2}
(D) \frac{1}{{52}}
Ans. (A)
Sol. P(E) =\frac{2}{{52}} = \frac{1}{{26}} (favourable cases are 2).
38. Three unbiased coins are tossed. What is the probability of getting exactly 3 heads :
(A) \frac{1}{4}
(B) \frac{3}{8}
(C)\frac{1}{8}
(D) \frac{1}{2}
Ans. (C)
Sol. P(E) =\frac{1}{8} (favourable cases is 1 HHH).
39. There are 200 tickets in a bag numbered 1 to 200. A ticket is drawn randomly from it. Find the probability that the number on ticket drawn is divisible by 4 and 3.
(A)\frac{2}{5}
(B)\frac{2}{{25}}
(C)\frac{{67}}{{200}}
(D) None of these
Ans. (B)
Sol. P(E) = \frac{{16}}{{200}} = \frac{2}{{25}} (favourable cases are 16).
40. Three unbiased coins are tossed. What is the probability of getting at most 2 heads ?:
(A) \frac{1}{4}
(B) \frac{3}{8}
(C) \frac{7}{8}
(D) \frac{1}{2}
Ans. (C)
Sol. P(E) =\frac{7}{8} .
41. From a normal pack of cards, a card is drawn at random. The probability of getting a jack or a king is
(A) \frac{2}{{52}}
(B) \frac{1}{{52}}
(C) \frac{2}{{13}}
(D) None of these
Ans. (C)
Sol. P(E) = \frac{8}{{52}} = \frac{2}{{13}}
(Favourable cases = 8) (4 Jack, 4 King)
42. A bag contains 3 white and 5 red balls. If a ball is drawn at random, the probability that the drawn ball is red is
(A) \frac{3}{8}
(B) \frac{5}{8}
(C) \frac{3}{{15}}
(D) \frac{5}{{15}}
Ans. (B)
Sol. P(E) =\frac{5}{{5 + 3}} = \frac{5}{8}
43. A card is drawn from a packet of 100 cards numbered 1 to 100. The probability of drawing a number which is a perfect square is
(A) \frac{1}{{10}}
(B) \frac{9}{{100}}
(C) \frac{1}{{100}}
(D) \frac{2}{{100}}
Ans. (A)
Sol. There are total 10 perfect squares from 1 to 100.
So, P(E) =\frac{{10}}{{100}} = \frac{1}{{10}}
44. If a coin is tossed twice, the probability of getting at least one head is
(A) \frac{1}{2}
(B) \frac{1}{4}
(C) \frac{3}{4}
(D) None of these
Ans. (C)
Sol. Total outcomes = 4(HH, HT, TH, TT)
P(E) = \frac{3}{4}
45. The probability of getting a number greater than 2 or an even number in a single throw of a fair die is
(A) \frac{1}{3}
(B) \frac{2}{3}
(C) \frac{5}{6}
(D) None of these
Ans. (C)
Sol. Favourable outcomes = 2, 3, 4, 5, 6
So, P(E) = \frac{5}{6}
46. The probability that in a family of 3 children, there will be at least one boy is
(A) \frac{7}{8}
(B) \frac{1}{8}
(C) \frac{4}{8}
(D) \frac{6}{8}
Ans. (A)
Sol. There are total 8 cases, in a family of 3 childrens
⇒ P(E) = \frac{7}{8}
47. In a simultaneously throw of three dices, find the probability of getting a triplet.
(A)\frac{1}{{36}}
(B) \frac{1}{{18}}
(C)\frac{1}{{216}}
(D) None of these
Ans. (A)
Sol. P(E) =\frac{6}{{216}} = \frac{1}{{36}}
48. The probability of guessing the correct answer to a question is\frac{x}{{12}} , if the probability of not guessing the correct answer to this question is \frac{2}{3}, then x =
(A)\frac{1}{3}
(B) 4
(C) 1
(D) 0
Ans. (B)
Sol. ⇒ \frac{2}{3} + \frac{x}{{12}}= 1
⇒ \frac{x}{{12}} =1-\frac{2}{3} = \frac{1}{3}
⇒ x = 4
49. One letter is chosen randomly from the letters of the word ‘MATHEMATICS’ what is the probability that the chosen letter is a consonant :
(A) \frac{4}{{11}}
(B) \frac{5}{{11}}
(C) \frac{7}{{11}}
(D) \frac{3}{{11}}
Ans. (C)
Sol. Total letters = 11
number of consonants = 7
So, P(E) = \frac{7}{{11}}
50. A book containing 100 pages is open at random. The probability that a doublet page is found, is :
(A)\frac{{90}}{{100}}
(B)\frac{9}{{100}}
(C) \frac{{99}}{{100}}
(D)\frac{{10}}{{100}}
Ans. (B)
Sol. Favourable cases = 9 (11, 22, 33, 44………….. 99)
P(E) =\frac{9}{{100}}
