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# Test: 35 Year JEE Previous Year Questions: Probability

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## 28 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: 35 Year JEE Previous Year Questions: Probability

Test: 35 Year JEE Previous Year Questions: Probability for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The Test: 35 Year JEE Previous Year Questions: Probability questions and answers have been prepared according to the JEE exam syllabus.The Test: 35 Year JEE Previous Year Questions: Probability MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: 35 Year JEE Previous Year Questions: Probability below.
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Test: 35 Year JEE Previous Year Questions: Probability - Question 1

### A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is and  . Probability that the problem is solved is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 1

Test: 35 Year JEE Previous Year Questions: Probability - Question 2

### A and B are events such that P(A ∪ B)=3/4, P(A ∩ B)=1/4, is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 2

P (A ∪ B) = P (A) + P (B) – P (A ∩ B);

Now,

Test: 35 Year JEE Previous Year Questions: Probability - Question 3

### A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 3

The event follows binomial distribution with
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.

Test: 35 Year JEE Previous Year Questions: Probability - Question 4

The mean and variance of  a random variable X having binomial distribution are 4 and 2 respectively, then P (X = 1) is
[2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 4

Test: 35 Year JEE Previous Year Questions: Probability - Question 5

Events A, B, C are mutually exclusive events such that  and  The set of possible values of x are in the interval. [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 5

and

∵ For any event E , 0 ≤ P(E )≤1

and

⇒ -1≤ 3x ≤ 2,-3 ≤ x ≤1 and -1 ≤ 2x≤ 1

Also for mutually exclusive events A, B, C,
P(A∪ B ∪C) = P(A) + P(B)+ P(C)

Considering all inequations, we get

Test: 35 Year JEE Previous Year Questions: Probability - Question 6

Five  horses  are  in a race.  Mr. A  selects two of  the horses  at random and  bets on them. The  probability that Mr. A selected the winning horse is [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 6

Let 5 horses are H1, H2, H3, H4 and H5. Selected pair of horses will be one of the 10 pairs (i.e.; 5C2 ): H1 H2, H1 H3, H1 H4, H1 H5, H2H3, H2 H4, H2 H5, H3 H4, H3 H5
and H4 H5.
Any horse can win the race in 4 ways.
For example : Horses H2 win the race in 4 ways H1 H2, H2H3, H2H4 and H2H5.
Hence required probability =

Test: 35 Year JEE Previous Year Questions: Probability - Question 7

The probability that A speaks truth is ,  while the probability for B is .
The probability that they contradicteach oth er when asked to speak on a fact is [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 7

A an d B will contradict each other if one speaks truth and other false . So , the required

Probability =

Test: 35 Year JEE Previous Year Questions: Probability - Question 8

A random variable X has the probability distribution:  For the events E = {X is a prime number } and F = {X< 4}, the P( E ∪ F) is [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 8

P(E) = P ( 2 or 3 or 5 or 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62
P( F ) =P(1 or2 or3) = 0.15 + 0.23 + 0.12 = 0.50 P( E ∩ F )=P(2 or 3) = 0.23 + 0.12 = 0.35
∴ P(EUF ) = P(E) + P(F ) - P(E∩ F )
= 0.62 + 0.50 - 0.35 = 0.77

Test: 35 Year JEE Previous Year Questions: Probability - Question 9

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 9

mean = np = 4 and variance = npq = 2

and n = 8

∴ P (2 success)

Test: 35 Year JEE Previous Year Questions: Probability - Question 10

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 10

For a particular house being selected Probability =

P (all the persons apply for the same house)

Test: 35 Year JEE Previous Year Questions: Probability - Question 11

A random variable X has Poisson distribution with mean 2.
Then P (X > 1.5) equals [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 11

According to Poission distribution, prob. of getting k successes is

=1- P(x = 0) - P(x=1)

Test: 35 Year JEE Previous Year Questions: Probability - Question 12

Let A and B be two events such that    and  where stands forcomplement of event A. Then events A and B are

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 12

and

Also ⇒ P(A ∪ B) = P(A) + P(B) - P(A∩ B)

Hence  A and B are independent but not equally likely.

Test: 35 Year JEE Previous Year Questions: Probability - Question 13

At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone  calls during 10 minute time intervals.
The probability that there is at the most one phone call during a 10-minute time period is [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 13

P (at most 1 phone call)

= P(X ≤ 1) = P(X = 0) + P(X= 1)

Test: 35 Year JEE Previous Year Questions: Probability - Question 14

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 14

Let A be the event that Ist airplane hit the target and B be the event that IInd airplane hit the target

Hence, the corresponding probabilities are:

P(A) = 0.3, P(A’) = 0.7

P(B) = 0.2, P(B’) = 0.8

Thus, the required probability

= P(A’)P(B) + P(A’)P(B’)P(A’)P(B) +…

=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)

= 0.14[1+(0.56)+(0.56)2]

On solving the above expression, we get

= 0.14/0.44

=7/22

=0.32

Test: 35 Year JEE Previous Year Questions: Probability - Question 15

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 15

A pair of fair dice is thrown, the sample space S = (1, 1), (1, 2) (1, 3) .... = 36
Possibility of getting 9 are (5, 4), (4, 5), (6, 3), (3, 6)
∴  Probability of getting score 9 in a single throw

∴ Probability of getting score 9 exactly twice

Test: 35 Year JEE Previous Year Questions: Probability - Question 16

It is given that the events A and B are such that   and Then P(B) is [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 16

P(A) = 1/4, P(A/B)  P(B/A) = 2/3
By conditional probability,

P(A ∩ B) =  P(A) P(B/A) = P(B)P(A/B)

Test: 35 Year JEE Previous Year Questions: Probability - Question 17

A die  is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A∪B) is [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 17

A ≡ number is greater than

B ≡ number is less than 5 ⇒

A ∩B≡ number is greater than 3 but less than 5.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Test: 35 Year JEE Previous Year Questions: Probability - Question 18

In a binomial distribution    if the probability of at least one success is greater than or equal to    then n is greater than: [2009]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 18

We have

Taking log to the base 3/4, on both sides, we get

Test: 35 Year JEE Previous Year Questions: Probability - Question 19

One ticket is selected at random from 50 tickets numbered 00,01,02,...,49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals: [2009]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 19

Let A ≡ Sum of the digits is 8
B ≡ Product of the digits is 0
Then  A = {08, 17, 26, 35, 44}
B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40,}
A ∩B = { 08 }

Test: 35 Year JEE Previous Year Questions: Probability - Question 20

Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ...20}. [2010]

Statement -1: The probability that the chosen numbers when arranged in some order will form an AP is

Statement -2 : If the four chosen numbers form an AP, then the set of all possible values of common difference is (±1, ±2, ±3, ±4,±5) .

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 20

n(S) = 20C4 Statement-1:
common difference is 1; total number of cases = 17
common difference is 2; total number of cases = 14
common difference is 3; total number of cases = 11
common difference is 4; total number of cases = 8
common difference is 5; total number of cases = 5
common difference is 6; total number of cases = 2

Prob. =

Statement -2 is false, because common difference can be 6 also.

Test: 35 Year JEE Previous Year Questions: Probability - Question 21

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 21

n( S) = 9C3 , n( E) = 3C1 x 4C1 x 2C1

Probability =

Test: 35 Year JEE Previous Year Questions: Probability - Question 22

Con sid er 5 in dependen t Ber n oull i’s tr i als each wi th probability of success p. If the probability of at least one failure is greater than or equal to   then p lies in the interval  [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 22

p (at least one failure)

⇒ 1 – p (no failure)

But p ≥ 0,

Hence p lies in the interval

Test: 35 Year JEE Previous Year Questions: Probability - Question 23

If C and D are two events such that C ⊂ D and P(D) ≠ 0, then the correct statement among the following is     [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 23

In this case,

Where, 0 ≤ P(D) ≤1 , hence

Test: 35 Year JEE Previous Year Questions: Probability - Question 24

Three numbers are chosen at random without replacement from {1,2,3,..8}. The probability that their minimum is 3, given that their maximum is 6, is : [2012]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 24

Given sample space = {1,2,3,.....,8}
Let Event A : Maximum of three numbers is 6.
B : Minimum of three numbers is 3.
This is the case of conditional probability
We have to find P (minimum) is 3 when it is given that P (maximum) is 6.

Test: 35 Year JEE Previous Year Questions: Probability - Question 25

A multiple choice examination has 5 questions. Each question has three alternative answers of which  exactly one is correct.
The probability that a student will get 4 or more correct answers just by guessing is: [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 25

By using Binomial distribution

Required probability =

Test: 35 Year JEE Previous Year Questions: Probability - Question 26

Let A and B be two events such that   where stands for the complement of the event A. Then the events A and B are [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 26

Given

We know P(A∪ B) = P(A)+ P(B)- P(A∩ B)

∵ P( A) ≠ P(B) so they are not equally likely.

Also

So A & B are independent.

Test: 35 Year JEE Previous Year Questions: Probability - Question 27

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is : [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 27

Note:- The question should state ‘3 different’ boxes instead of ‘3 identical boxes’ and one particular box has 3 balls. Then the solution would be:

Required probability =

Test: 35 Year JEE Previous Year Questions: Probability - Question 28

Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Probability - Question 28

And P(E1 ∩ E2 ∩ E3) = 0 ≠ P (E1) . P(E2) . P(E3)

⇒ E1, E2, E3 are not independent.

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