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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]
A and B are events such that P(A ∪ B)=3/4, P(A ∩ B)=1/4, is [2002]
P (A ∪ B) = P (A) + P (B) – P (A ∩ B);
Now,
A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]
The event follows binomial distribution with
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.
The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P (X = 1) is
[2003]
Events A, B, C are mutually exclusive events such that and The set of possible values of x are in the interval. [2003]
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
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]
Let 5 horses are H_{1}, H_{2}, H_{3}, H_{4 }and H_{5}. Selected pair of horses will be one of the 10 pairs (i.e.; ^{5}C_{2} ): H_{1} H_{2}, H_{1} H_{3}, H_{1 }H_{4}, H_{1} H_{5}, H_{2}H_{3}, H_{2} H_{4}, H_{2} H_{5}, H_{3} H_{4}, H_{3} H_{5}
and H_{4} H_{5}.
Any horse can win the race in 4 ways.
For example : Horses H_{2} win the race in 4 ways H_{1} H_{2}, H_{2}H_{3}, H_{2}H_{4} and H_{2}H_{5}.
Hence required probability =
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]
A an d B will contradict each other if one speaks truth and other false . So , the required
Probability =
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]
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
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is [2004]
mean = np = 4 and variance = npq = 2
and n = 8
∴ P (2 success)
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]
For a particular house being selected Probability =
P (all the persons apply for the same house)
A random variable X has Poisson distribution with mean 2.
Then P (X > 1.5) equals [2005]
According to Poission distribution, prob. of getting k successes is
=1 P(x = 0)  P(x=1)
Let A and B be two events such that and where stands forcomplement of event A. Then events A and B are
and
Also ⇒ P(A ∪ B) = P(A) + P(B)  P(A∩ B)
Hence A and B are independent but not equally likely.
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 10minute time period is [2006]
P (at most 1 phone call)
= P(X ≤ 1) = P(X = 0) + P(X= 1)
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]
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
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007]
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
It is given that the events A and B are such that and Then P(B) is [2008]
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)
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]
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)
In a binomial distribution if the probability of at least one success is greater than or equal to then n is greater than: [2009]
We have
Taking log to the base 3/4, on both sides, we get
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]
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 }
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) .
n(S) = ^{20}C_{4} Statement1:
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.
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]
n( S) = ^{9}C_{3} , n( E) = ^{3}C_{1} x ^{4}C_{1} x ^{2}C_{1}
Probability =
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]
p (at least one failure)
⇒ 1 – p (no failure)
But p ≥ 0,
Hence p lies in the interval
If C and D are two events such that C ⊂ D and P(D) ≠ 0, then the correct statement among the following is [2011]
In this case,
Where, 0 ≤ P(D) ≤1 , hence
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]
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.
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]
p = P (correct answer), q = P (wrong answer)
By using Binomial distribution
Required probability =
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]
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.
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]
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 =
Let two fair sixfaced dice A and B be thrown simultaneously. If E_{1} is the event that die A shows up four, E_{2 }is the event that die B shows up two and E_{3} 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]
And P(E_{1} ∩ E_{2} ∩ E_{3}) = 0 ≠ P (E_{1}) . P(E_{2}) . P(E_{3})
⇒ E_{1}, E_{2}, E_{3} are not independent.
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