Practice question On probability For IIT-Jee, Airforce and NDA

Probability

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Question no. 1

A coin is tossed three times. If events A and b are defined as A = Two heads come , B = last should be head. Then A and B are

looks_one Independent

looks_two dependent

looks_3 both

looks_4 none of these

Option looks_two dependent

very soon 16

Question no. 2

If one ball is drawn at random from each of three boxes containing 3 white and 1 black , 2 white and 2 black , 1 white and 3 black balls , then tge probability that 2 white and 1 black balls will be drawn is

looks_one

looks_two

looks_3

looks_4

Option looks_one

ver soon 2

Question no. 3

A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_one

solution :
Each suit has 13 cards and there are 4 suit.
The probability that all the four cards drawn are of the same suit = 4*(13*12*11*10)/(52*51*50*49) = 44/(85*49)

Question no. 4

A bag contains 9 balls , two of which are red , three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is

looks_one

looks_two

looks_3

looks_4

Option looks_one

Solution :
Probability of having 3 blue balls = (3*2*1)/(9*8*7)= 1/84
Probability of having 3 black balls = (4*3*2)/(9*8*7)= 4/84
Probability that they are of the same colour = 1/84 + 4/84 = 5/84

Question no. 5

Two dice are thrown simultaneously. The probability of getting a pair of aces is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_one

soltion :
Total number of combination = 36 probability of getting a pair of aces (1,1) = 1/36

Question no. 6

A bag contains 5 black balls , 4 white balls and 3 red balls. If a bll is selected randomly , the probability that it is black or red ball is

looks_one

looks_two

looks_3

looks_4

Option looks_4

solution :
probability that it is black or red ball = 5/12 + 3/12 = 8/12 = 2/3

Question no. 7

Out of 30 consecutive integers , 2 are chosen at random. The probability that their sum is odd , is

looks_one

looks_two

looks_3

looks_4

Option looks_3

Solution :
for sum to be odd , One number should be positive and another number should be even
probability that their sum is odd = (first odd ) * (second even) + (first even) * (second odd)
= (15/30) * (15*29) + (15/30) * (15*29) = 15/29

Question no. 8

Three integers are chosen at random from the first 20 integers. The probability that their product is even is

looks_one

looks_two

looks_3

looks_4

Option looks_3

Solution :
probability of (having odd product (all the numbers should be odd)) = (10*9*8)/(20*19*18) = 2/19
The probability that their product is even = 1 - 2/19 = 17/19

Question no. 9

A speaks truth in 75% cases and B speaks truth in 80% cases. Probability that they contradict each other in a statement , is

looks_one

looks_two

looks_3

looks_4

Option looks_one

Solution :
condition or Contradictory = A(speaks false ) + B(speaks true ) or B(speaks false ) + A(speaks true )
probability(A(speaks false ) + B(speaks true ) ) = (1/4)*(4/5) =4/20
probability(B(speaks false ) + A(speaks true ) ) = (3/4)*(1/5) =3/20
Probability that they contradict each other = 4/20 + 3/20 = 7/20

Question no. 10

A person write 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is

looks_one

looks_two

looks_3

looks_4

Option looks_4

Solution :
Total number of combination = 4! = 24
Only one way to get the letter in correct envelope
there are 23 ways to get incorrect way of letter put in envlopeee
probability that all letters are not placed in the right envelopes = 23/24

Question no. 11

The probability that a leap year will have 53 Fridays or 53 Saturdays is

looks_one

looks_two

looks_3

looks_4

Option looks_two

Solution :
Leap year has 52 weeks and 2 days , so 2 days can be
Saturday + Sunday
Sunday + Monday
Monday + tuesday
tuesday + Wednesday
Wednesday + thursday
thursday + friday
friday + saturday
Probability(friday) = 2/7
Probability(saturday) = 2/7
Probability(friday and satudray) = 1/7
Probability(friday or satudray) = Probability(friday) + Probability(saturday) - Probability(friday and satudray) = 2/7 +2/7 - 1/7 = 3/7

Question no. 12

Three faces of an ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled 3 times. The probability that yellow red and blue face appear in the first second and third thrown respectively, is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_one

Solution :
The probability that yellow red and blue face appear in the first second and third thrown respectively = (3/6)* (2/6) *(1/6) = 1/36

Question no. 13

India play two matches each with West Indies and Australia. In any match the probabilities of india getting 0, 1 and 2 points are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India geting at least 7 points is

looks_one 0.06875

looks_two 1/16

looks_3 0.1125

looks_4 None of these

Option looks_one 0.06875

Solution :
Maximum pont = 8
Probability(Getting 7 point ) ( 2 points must be scored in 3 matches ) = 0.05 * (0.5)³ = 0.00625
Probability(Getting 8 point ) ( 2 points must be scored in 4 matches ) = (0.5)⁴ = 0.0625
Probability(geting at least 7 points) = 0.00625 + 0.0625 = 0.06875

Question no. 14

The probabilities of a student getting 1, Il and il division in an examination are 1/10 , 3/5 and 1/4 respectively. The probability that the student fails in the examination is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_4 None of these

Solution :
Probability of fails in examination = 1 - 1/10 - 3/5 - 1/4 = 9/20

Question no. 15

A and B are two events such as P(A) = .25 and P(B) = 0.50. The probability of both happening together is 0.14 .The probability of both A and B not happening is

looks_one 0.39

looks_two 0.25

looks_3 0.11

looks_4 None of these

Option looks_one 0.39

Solution :
P(A) = 0.25
P(B) = 0.5
P(A.B) = 0.14
P(A+B) = P(A) + P(B) - P(A.B) = 0.5 + 0.25 - 0.14
P(A+B) = 0.61
P'(A+B) = 1 - P(A+B) = 1 - 0.61 = 0.39

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