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SOLVED PAPER : SEPTEMBER 2025 (SUGGESTED ANSWERS ) P.1
SOLVED PAPER : JANUARY 2026 (SUGGESTED ANSWERS ) P.29
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PROBABILITY
Experiment: The performance which produces the certain results is called Experiment.
Random Experiment: If the results of the experiment depend on chance only, then the experiment is called Random Experiment. Example: Tossing a coin; Throwing a die; drawing a card from well - shuffled pack of 52 cards etc.
Sample Space: The set of all possible distinct outcomes of a random experiment is called SAMPLE SPACE (or Event Space). It is denoted by capital letter “S”.
Example 1: A coin is tossed at random then S = {H, T} n(S) = 2
Either Head (H) or Tail (T) can occur on upper face of the coin.
Example 2: A die is thrown at random then S = {1, 2, 3, 4, 5, 6} n(S) = 6
A die has 6 faces with face number 1, 2, 3, 4, 5, 6. One of them can occur at a time.
Example 3: Two coins are tossed together then S = {HH, HT; TH; TT} i.e. If both coins are tossed together then heads on both coins or Tails on both coins or Head on one coin and Tail on another one coin can occur. Another way to find Sample Space.
Total Sample Space “S” = cross - product of individual Sample - Space.
S = {H, T} {H, T}
= {H H, HT, TH, TT}
n(S) = n(S1)× n(S2) = 2 2 = 4
Tricks: For Coins n(S) = 2 No. of coins tossed together
Example 1: For 2 coins
n(S) = 22 = 4
Example 3: For 3 coins tossed together
n(S) = 23 = 8
For Dice
23.2
PROBABILITY
Example 2: If two dice are thrown together then, S = {1, 2, 3, 4, 5, 6} {1, 2, 3, 4, 5, 6} = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6); (2, 1), (2, 2), ................................; (2, 3); (3, 1), (3, 2);.................................;(3, 6); (4, 1), (4, 2);.................................;(4, 6); (5, 1), (5, 2);.................................;(5, 6); (6, 1), (6, 2);.................................;(6, 6)}
36. 6 6 ) ( ). ( ) ( 2 1 s n s n s n
Tricks: For Dice
n(S) = 6 No. of dice Thrown together .
PAST EXAM QUESTIONS WITH SOLUTIONS (MEMORY BASED)
Q.1. There are 6 positive and 8 negative numbers. Four number are selected at random without replacement and multiplied. Find the probability that the product is positive.
( a ) 1001 420 ( b ) 1001 409
(c ) 1001 70 ( d ) 1001 505 [June 2015]
Solution: (d) is correct.
Let 6 positive Nos. are 1, 2, 3, 4, 5, 6, and 8 negative Nos. are –1,–2, –3, ........, –8
Sample space
= n(s) = 14 4 14! C (4!)(10!) = 1413121110! 432110! = 1001
Let Event = E = such that product of them is positive
n(E) = 6688 4224 CCCC = (All + ve) (Two +ve & two -ve) + (All 4 are -ve)
= 15 + 15 × 28 + 70 = 505
P(E) = 1001 505 ) ( ) ( S n E n
Q.2. 1212 P(A)3/8;P(A)2/3;PAA=1/4 then A12 and A will be
( a )Mutually exclusive & independent ( b )Exclusive but not independent
(c )Independent but not exclusive
( d )None [June 2015]
Solution: (c) is correct. 4 1 ) A A( P 2 1 (given) 0 2 1 A & A are not Mutually Exclusive Events
2212 321 P(AA)P(A)P(A) 834
Clearly z AandA12 are Independent Events
Q.3. The sum of two numbers obtained in a single throw of two dice is ‘S’. Then the probability of ‘s’ will be maximum when ‘S’ =
( a )5( b )7
(c )6( d )8 [June 2015]
Solution: (b) is correct.
S = Sum of face values of two due.
S = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
n(s) = 6 it is maximum if sum of their face is 7.
Q.4. When an unbiased dice is rolled, find the odds in favour of getting of multiple of 3.
( a )1/6( b )1/4
(c )1/2( d )1/3 [Dec. 2015]
Solution: (c) is correct.
S = {1, 2, 3, 4, 5, 6}
Let E = {3,6}; E1 = {1,2,3,4,5,6}-{3,6} = {1, 2, 4, 5} n(E) = 2; n(E1) = 4
Odds in favour of Event
E = 1 ()21 ()42 n n
Q.5. Three coins are rolled, what is the probability of getting exactly two heads:
( a )1/8( b )3/8
(c )7/8( d )5/8 [Dec. 2015]
Solution: (b) is correct.
Given 3 n P = Prob. (head) in 1 trial = 2 1 2 1 2 1 1 1 p q 2 321 2 113 23 228 Cpq
Q.6. If a random sample of 500 Oranges produces 25 rotten oranges. Then the estimate of the proportion of rotten oranges in the sample is:
( a )0.01( b )0.05
(c )0.028( d )0.0593 [Dec. 2015]
Solution: (b) is correct. 0.05 500 25
23.4
PROBABILITY
Q.7. Two letter are drawn at random from word “HOME” find the probability that there is no vowel.
( a )5/6( b )1/6
(c )1/3( d )None
[Dec. 2015]
Solution: (b) is correct. 4 2 ()6.nSC
E = Event of no vowel = {H; M}
(E) n 1 2 2C () 1 () ()6 n nS
Q.8. A bag contains 15 one rupee coins, 25 two rupee coins and 10 five rupee coins. If a coin is selected at random from the bag, then the probability of not selecting a one rupee coin is:
( a )0.30( b )0.70
(c )0.25( d )0.20
[Dec. 2015]
Solution: (b) is correct. 0.70 50 10 25
Q.9. If 235 P(A),P(B),P(AB), 356 then (A/B) is:
( a ) 12 7 ( b ) 12 5
(c ) 4 1 ( d ) 2 1
[June 2016]
Solution: (b) is correct.
Q.10. Two dice are tossed what is the probability that the total is divisible by 3 or 4:
( a ) 36 20 ( b ) 36 21
(c ) 36 14 ( d )None [June 2016]
Solution: (a) is correct. ()36nS
A = Event of getting Nos. such that the sum of their face Nos. is divisible by 3 = (sum 3sum 6sum 9sum 12)
= {(1, 2), (2, 1); (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} (A)12n
B = Event of getting Nos. such that the sum of their face Nos. is divisible by 4 = (sum 4sum 8sum 12)
= {(1, 3), (2, 2); (3, 1); (2, 6), (3, 5), (4, 4); (5, 3); (6, 2); (6, 6)} (B)9n
PROBABILITY 23.5
B A = Divisible by LCM of 3 & 4 = 12 = sum 12 = {(6, 6)} (AB)1n (A) (B) (AB) P (AB) (S) nnn n 36 1 9 12 9 5 36 20
(a) is correct.
Q.11. If 2 dice are rolled simultaneously then the probability that their sum is neither 3 nor 6 is
( a )0.5( b )0.75 (c )0.25( d )0.80 [June 2016]
Solution: (d) is correct.
A= event that sum is 3 = {(1, 2); (2, 1)}
2 A n
B= event that sum is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
5 B n
Q.12. In a game, cards are thoroughly shuffled and distributed equally among four players. What is the probability that a specific player gets all the four kings?
( a ) 5248 413 52 11 CC C ( b ) 448 49 52 13 CC C
(c ) 1339 99 52 13 CC C ( d ) 439 49 52 13 CC C [June 2016]
Solution: (b) is correct. Each candidate will get 13 cards. 52 13 n(S) C let E = Event of getting all 4 kings and rest 9 cards by a player out of 48. 448 n(E)CC49 448 49 52 13 CC n(E) P(E) n(S) C
Q.13. A bag contains 4 red and 5 black balls. Another bag contains 5 red, 3 black balls. If one ball is drawn at random from each bag. Then the probability that one red and one black ball drawn is ______.
( a ) 72 12 ( b ) 72 25
(c ) 72 37 ( d ) 72 13 [June 2016]
Solution: (c) is correct.
Prob. (1 red and 1 black balls) = P (R1) . P (B2) + P (B1) . P (R2)
Where R1 = event of getting red ball from 1st bag.
23.6
PROBABILITY
R2 = ””””2nd ”
B1 =””Black ball from 1st bag. B2 =”””2 nd ”
Prob. = 1 8 1 5 1 9 1 5 1 8 1 3 1 9 1 4 C C C C C C C C = 72 37 72 25 12 8 5 9 5 8 3 9 4
Q.14. In a discrete random variable follows uniform distribution and assumes only the value 8, 9, 11, 15, 18, 20. Then P(X 15 ) is ______.
( a )1/2( b )1/3
(c )2/3( d )2/7 [June 2016]
Solution: (c) is correct.
E = Event of Nos. 15{8,9,11,15} 3 2 6 4 (S) ) E ( 15 X P n n
Q.15. A bag contains 6 green and 5 red balls. One ball is drawn at random. The probability of getting a red ball is?
Solution: (a) is correct.
n(S) = 11 1C = 11
Let E = Event of getting a red ball. n (E) = 5 1C = 5 P (E) = 5 11
Q.16. If two events A, B P(A) = 1 2 ; P(B) = 1 3 and P(A B) = 2 3 , then find P(A B)? ( a ) 4 1 ( b ) 6 1 (c ) 3 2 ( d ) 2 1 [Dec. 2016]
Solution: (b) is correct. P(AB) = P(A) + P(B) – P (A B) = 1 2 + 1 3 –2 3 = 6 1
( a ) 11 5 ( b ) 11 6 (c ) 6 5 ( d )None [Dec. 2016]
Q.17. If P(A),P(B),P(AB)231 384 , then the events A & B are _______.
( a )Independent and mutually exclusive
( b )Independent but not mutually exclusive
(c )Mutually exclusive but not independent
( d )Neither Independent nor exclusive [Dec. 2016]
Solution: (b) is correct.
PROBABILITY
Since, P(AB) 0, So A & B are not mutually exclusive events.
Now, P(AB) = P(A)×P(B) = 231 384
Hence, Events A & B are Independent events. So, (b) is correct.
23.7
Q.18. The probability of getting atleast one 6 from 3 throws of a perfect die is
) 6
Solution: (d) is correct.
n = 3 (Trials)
P = Prob. of getting 6 in 1 trial = 1 6
q = 1 – p = 1 –1 6 = 5 6
P11P1 xx = 303 0 1P01CPq x = 11.115533 66
Q.19. For any two events A and B
( a )P(A-B) = P(A)-P(B)
( b )P(A-B) = P(A) - P(A B)
(c )P(A-B) = P(B) - P(A B)
[June 2017]
( d )P(B-A) = P(B) + P(A B) [June 2017]
Solution: (b) is correct.
Q.20. If PA,PB,PAB211 3412 then B P ____ A
( a ) 1 8 ( b ) 7 8 (c ) 1 3 ( d ) 2 5 [June 2017]
Solution: (a) is correct.
23.8
PROBABILITY PAB B P A PA = 1 131 12 2 1228 3
Q.21. For the events A & B if PA,PB11 23 and 1 PAB 4 then A P B
( a )1/2( b )1/6( c )2/3( d )3/4 [Dec. 2017]
Solution: (d) is correct. 1 PAB 4
Q.22. If A & B are two mutually exclusive events such that: , 5 2 A P , 3 2 B A P then P (B):
( a )4/15( b )4/9 (c )5/9( d )7/15 [Dec. 2017]
Solution: (a) is correct PABPAPB
[ A & B are two mutually exclusive events] 22 PB 35 or; 22106 PB 3515 4 15 i ii iii ( a ) 35 24 , 35 10 , 35 1 ( b ) 35 14 , 35 7 , 35 24 (c ) 35 11 , 35 24 , 35 3 ( d ) 35 20 , 35 6 , 35 24 [Dec. 2017]
Solution: (a) is correct. Let A and B are events of selection of brother and sister respectively. Both events are independent. (i) P(Both selected)
(ii) P(Only one will be selected]
(iii) P(None of them will be selected) 11 11 PAPB11 75 6424 7535
Q.24. If 4 letters are put randomly among the 4 envelopes then the probability that all are not put in correct envelopes:
( a )1/24( b )1
(c )23/24( d )9/24 [June 2018]
Solution: (c) is correct.
n(s) = 4! = 24
Let E = Events of putting letter in right envelop.
n(E) = 1×1×1×1 = 1 nE 1 pE nS24 1 23 1 pE1 2424
Q.25. Two broad divisions of probability are:
( a )Subjective probability and objective probability
( b )Deductive probability and mathematical probability
(c )Statistical probability and mathematical probability
( d )None of these [May 2018]
Solution: (a) is correct. Two broad divisions of Probability are:
(i)Subjective Probability
(ii) Objective Probability
Q.26. The term “chance” and probability are synonyms:
( a )True( b )False
(c )Both( d )None [May 2018]
Solution: (a) is correct.
Q.27. The theorem of Compound Probability states that for any two events A and B:
( a ) PABPAPB/A
( b ) PABPAPB/A
(c ) PABPAPB
( d ) PABPAPBPAB
[CA (F) May 2018]
23.10
Solution:
PROBABILITY
(a) The theorem of Compound Probability states that for two events A and B PABPAPB/A
Q.28. Variance of random variable x is given by:
( a ) 2 EX
( b ) 2 EXEX
(c ) 2 EX
( d )(a) or (b) [May 2018]
Solution:
(d) Variance of a random variable x is given by 2 VxEX or 2 VxEXEX
Note: = E(X)
Q.29. What is the probability of having at least one ‘six’ appear in 3 throws of a perfect die?
( a )5/6( b ) 3 5/6
(c ) 3 11/6 ( d ) 3 15/6 [May 2018]
Solution : (d) For a die Probability of getting Six 1 PAp 6 15 PA1q 66 Here n = 3
P(getting at least ‘1’ Six) = P(X 1) = 1PX1 = 1PX0
Q.30. Sum of all probabilities of mutually exclusive and exhaustive events is equal to:
( a )0( b )1/2
(c )1/4( d )1 [Nov. 2018]
Solution: (d) is correct. If events are mutually exclusive and exhaustive events then Sum of all probabilities = 1.
Q.31. If PA,PB,andPAB111 234 then PAB is equal to:
