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All questions of Probability and Statistics for Mathematics Exam
The probability that a man who is 85 yrs. old will die before attaining the age of 90 is 1 / 3. A1, A2, A3 and A4 are four persons who are 85 yrs. old. The probability that A1 will die before attaining the age of 90 and will be the first to die is- a)65/81
- b)13/81
- c)65/324
- d)13/108
Correct answer is option 'C'. Can you explain this answer?
The probability that a man who is 85 yrs. old will die before attaining the age of 90 is 1 / 3. A1, A2, A3 and A4 are four persons who are 85 yrs. old. The probability that A1 will die before attaining the age of 90 and will be the first to die is
a)
65/81
b)
13/81
c)
65/324
d)
13/108
|
Veda Institute answered |
Required probability = P[at least one person dies before 90 days) x P(first person to die is A1)
Subway trains on a certain line run every half hour between midnight and six in the morning. Find the probability that a person entering the station at a random time during this period will have to wait at least twenty minutes. - a)1/2
- b)2/3
- c)1/3
- d)1/6
Correct answer is option 'C'. Can you explain this answer?
Subway trains on a certain line run every half hour between midnight and six in the morning. Find the probability that a person entering the station at a random time during this period will have to wait at least twenty minutes.
a)
1/2
b)
2/3
c)
1/3
d)
1/6
|
Chirag Verma answered |
Waiting time of person will lie in the range (0,30) and in this range favourable portion is (20, 30), so the required probability is the ratio of the time interv l's length i.e. 
There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls is.- a)6
- b)7
- c)8
- d)9
Correct answer is option 'B'. Can you explain this answer?
There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls is.
a)
6
b)
7
c)
8
d)
9
|
|
Chirag Verma answered |
Possible configuration of balls to be selected seven possible cases are there.
A letter is taken at random from the letters of the word ‘STATISTICS ’ and an other letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that, they are the same letter is .- a)1/45
- b)13/90
- c)19/90
- d)5/8
Correct answer is option 'C'. Can you explain this answer?
A letter is taken at random from the letters of the word ‘STATISTICS ’ and an other letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that, they are the same letter is .
a)
1/45
b)
13/90
c)
19/90
d)
5/8
|
|
Chirag Verma answered |
The same letter can be either, S,T, I or A.
Probability required will be.
Probability required will be.
The mean of a class is 50 and variance is 9. Then what is the standard error of the marks?- a)5
- b)4
- c)3
- d)2
Correct answer is option 'C'. Can you explain this answer?
The mean of a class is 50 and variance is 9. Then what is the standard error of the marks?
a)
5
b)
4
c)
3
d)
2
|
|
Veda Institute answered |
Standard error = standard deviation
Let (n, p) and λ be the parameters of binomial and poisson distributions respectively. Consider the statements
P. The mean of the binomial distribution is np
Q. The standard deviation of the binomial distribution is np (1 - p)
R. The mean of the poisson distribution is λ
S The variance of the poisson distribution is λ
Which of the following group of statements is correct?- a)P,R,S
- b)P,Q,S
- c)P,Q,R
- d)P,Q,R,S
Correct answer is option 'A'. Can you explain this answer?
Let (n, p) and λ be the parameters of binomial and poisson distributions respectively. Consider the statements
P. The mean of the binomial distribution is np
Q. The standard deviation of the binomial distribution is np (1 - p)
R. The mean of the poisson distribution is λ
S The variance of the poisson distribution is λ
Which of the following group of statements is correct?
P. The mean of the binomial distribution is np
Q. The standard deviation of the binomial distribution is np (1 - p)
R. The mean of the poisson distribution is λ
S The variance of the poisson distribution is λ
Which of the following group of statements is correct?
a)
P,R,S
b)
P,Q,S
c)
P,Q,R
d)
P,Q,R,S
|
|
Charvi Verma answered |
R be two integers such that n > p > r > 0.
To prove that (n + p) choose r is equal to (n choose r) + (p choose r), we can use the definition of combinations.
According to the definition, (n choose r) represents the number of ways to choose r objects from a set of n objects, and (p choose r) represents the number of ways to choose r objects from a set of p objects.
Now, let's consider (n + p) choose r. This represents the number of ways to choose r objects from a set of (n + p) objects.
To count the number of ways to choose r objects from a set of (n + p) objects, we can divide the objects into two groups: n objects and p objects.
There are two cases to consider:
Case 1: We choose r objects only from the n objects group.
In this case, the number of ways to choose r objects from the n objects group is (n choose r), as we discussed earlier.
Case 2: We choose r objects only from the p objects group.
In this case, the number of ways to choose r objects from the p objects group is (p choose r).
Since these two cases are mutually exclusive (we cannot choose r objects from both groups simultaneously), we can add the number of ways from each case to get the total number of ways to choose r objects from the set of (n + p) objects.
Therefore, (n + p) choose r = (n choose r) + (p choose r).
This completes the proof.
To prove that (n + p) choose r is equal to (n choose r) + (p choose r), we can use the definition of combinations.
According to the definition, (n choose r) represents the number of ways to choose r objects from a set of n objects, and (p choose r) represents the number of ways to choose r objects from a set of p objects.
Now, let's consider (n + p) choose r. This represents the number of ways to choose r objects from a set of (n + p) objects.
To count the number of ways to choose r objects from a set of (n + p) objects, we can divide the objects into two groups: n objects and p objects.
There are two cases to consider:
Case 1: We choose r objects only from the n objects group.
In this case, the number of ways to choose r objects from the n objects group is (n choose r), as we discussed earlier.
Case 2: We choose r objects only from the p objects group.
In this case, the number of ways to choose r objects from the p objects group is (p choose r).
Since these two cases are mutually exclusive (we cannot choose r objects from both groups simultaneously), we can add the number of ways from each case to get the total number of ways to choose r objects from the set of (n + p) objects.
Therefore, (n + p) choose r = (n choose r) + (p choose r).
This completes the proof.
If A and B are two independent events in a given sample space and the probability that both A and B occur is 0.16 while the probability that neither occurs is 0.36 the probabilities P[A) and P[B) are respectively- a)0.6,0.6
- b)0.8,0.2
- c)0.4,0.4
- d)0.9,0.4
Correct answer is option 'C'. Can you explain this answer?
If A and B are two independent events in a given sample space and the probability that both A and B occur is 0.16 while the probability that neither occurs is 0.36 the probabilities P[A) and P[B) are respectively
a)
0.6,0.6
b)
0.8,0.2
c)
0.4,0.4
d)
0.9,0.4
|
|
Chirag Verma answered |
A and B throw a die in succession to win a bet with A starting first. Whoever throws '1' first wins an amount of Rs. 110. What are the respective expectations of A and B?- a)Rs. 70 and Rs. 40
- b)Rs. 60 and Rs. 50
- c)Rs. 75 and Rs. 35
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A and B throw a die in succession to win a bet with A starting first. Whoever throws '1' first wins an amount of Rs. 110. What are the respective expectations of A and B?
a)
Rs. 70 and Rs. 40
b)
Rs. 60 and Rs. 50
c)
Rs. 75 and Rs. 35
d)
none of these
|
|
Chirag Verma answered |
For a distribution 
variance = 100. The coefficient of variation is:- a)12.56
- b)14.236
- c)19.23
- d)21.02
Correct answer is option 'C'. Can you explain this answer?
For a distribution variance = 100. The coefficient of variation is:
variance = 100. The coefficient of variation is:a)
12.56
b)
14.236
c)
19.23
d)
21.02
|
Naveen Patel answered |
We know that coefficient of variation is equal to
(Standard deviation ÷ mean )×100
here standard deviation is 10
and mean is 52. then we will get 19.230
option c is correct
(Standard deviation ÷ mean )×100
here standard deviation is 10
and mean is 52. then we will get 19.230
option c is correct
A car is parked by owner amongst 25 cars in row, not at either end. On his return, he finds that exactly 15 places are still occupied. The probability that both the neighbouring parking places are empty is- a)91/276
- b)15/164
- c)15/92
- d)none of the above
Correct answer is option 'C'. Can you explain this answer?
A car is parked by owner amongst 25 cars in row, not at either end. On his return, he finds that exactly 15 places are still occupied. The probability that both the neighbouring parking places are empty is
a)
91/276
b)
15/164
c)
15/92
d)
none of the above
|
|
Chirag Verma answered |
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If E is the event that a number greater than or equal to 4 occurs on a single toss of the die, then P (E) is- a)1/2
- b)2/3
- c)4/9
- d)1/3
Correct answer is option 'C'. Can you explain this answer?
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If E is the event that a number greater than or equal to 4 occurs on a single toss of the die, then P (E) is
a)
1/2
b)
2/3
c)
4/9
d)
1/3
|
|
Chirag Verma answered |
Let probability of occurence of even number be k, then probability of occurence of odd number will be 2 k.
Suppose an interactive computer system is proposed for which it is estimated that the mean response time E(T) = 0.5 sec and standard deviation σ = 0.1 sec. Using Chebyshev's inequality, the probability P[|T-0.5| ≥ 0.25] is- a)0.16
- b)0.84
- c)0.25
- d)0.75
Correct answer is option 'A'. Can you explain this answer?
Suppose an interactive computer system is proposed for which it is estimated that the mean response time E(T) = 0.5 sec and standard deviation σ = 0.1 sec. Using Chebyshev's inequality, the probability P[|T-0.5| ≥ 0.25] is
a)
0.16
b)
0.84
c)
0.25
d)
0.75
|
|
Chirag Verma answered |
By chebvshev's inequality
A ship arrives at a port and 40 sailors on board go ashore for revelry. Later at night, the 40 sailors return to the ship and in their state of inebriation, each chooses a random cabin to sleep in. What is the expected number of sailors sleeping in their own cabin?- a)1/40
- b)1/4
- c)(1/40)10
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
A ship arrives at a port and 40 sailors on board go ashore for revelry. Later at night, the 40 sailors return to the ship and in their state of inebriation, each chooses a random cabin to sleep in. What is the expected number of sailors sleeping in their own cabin?
a)
1/40
b)
1/4
c)
(1/40)10
d)
None of these
|
|
Chirag Verma answered |
probability that the sailor returns in its own cabin is 1/40 If sailor is in his own cabin in the expected value head count will be increased by 1.
Suppose that the number of typographical errors on a single page of a certain book has a Poisson distribution with parameter λ = 1/2. In 600 pages book, the average number of errors in the book is- a)300
- b)150
- c)600
- d)393
Correct answer is option 'A'. Can you explain this answer?
Suppose that the number of typographical errors on a single page of a certain book has a Poisson distribution with parameter λ = 1/2. In 600 pages book, the average number of errors in the book is
a)
300
b)
150
c)
600
d)
393
|
|
Chirag Verma answered |
Mean or average of poisson distribution equals its parameter λ
A problem is given to three students A,B and C whose chances of solving it are 1/2 ,1/3 and 1/4 respectively. The probability that this problem will be solved, is :- a)1/24
- b)1/6
- c)2/3
- d)3/4
Correct answer is option 'D'. Can you explain this answer?
A problem is given to three students A,B and C whose chances of solving it are 1/2 ,1/3 and 1/4 respectively. The probability that this problem will be solved, is :
a)
1/24
b)
1/6
c)
2/3
d)
3/4
|
Devika Choudhary answered |
Understanding the Problem
In this scenario, we have three students A, B, and C with different probabilities of solving a problem:
- Student A: Probability of solving = 1/2
- Student B: Probability of solving = 1/3
- Student C: Probability of solving = 1/4
Calculating Individual Probabilities
To find the total probability that at least one of them solves the problem, we first calculate the probabilities of each student not solving it:
- Probability that A does NOT solve = 1 - 1/2 = 1/2
- Probability that B does NOT solve = 1 - 1/3 = 2/3
- Probability that C does NOT solve = 1 - 1/4 = 3/4
Finding the Combined Probability of Not Solving
Next, we multiply these individual probabilities together to find the probability that none of them solves the problem:
- Probability(None solve) = (1/2) * (2/3) * (3/4)
Calculating this gives:
- Probability(None solve) = (1/2) * (2/3) * (3/4) = 1/4
Calculating the Probability that the Problem is Solved
To find the probability that at least one student solves the problem, we subtract the probability that none solve from 1:
- Probability(At least one solves) = 1 - Probability(None solve)
- Probability(At least one solves) = 1 - 1/4 = 3/4
Conclusion
Thus, the probability that the problem will be solved by at least one of the students A, B, or C is:
- 3/4
This matches the correct answer option 'D'.
In this scenario, we have three students A, B, and C with different probabilities of solving a problem:
- Student A: Probability of solving = 1/2
- Student B: Probability of solving = 1/3
- Student C: Probability of solving = 1/4
Calculating Individual Probabilities
To find the total probability that at least one of them solves the problem, we first calculate the probabilities of each student not solving it:
- Probability that A does NOT solve = 1 - 1/2 = 1/2
- Probability that B does NOT solve = 1 - 1/3 = 2/3
- Probability that C does NOT solve = 1 - 1/4 = 3/4
Finding the Combined Probability of Not Solving
Next, we multiply these individual probabilities together to find the probability that none of them solves the problem:
- Probability(None solve) = (1/2) * (2/3) * (3/4)
Calculating this gives:
- Probability(None solve) = (1/2) * (2/3) * (3/4) = 1/4
Calculating the Probability that the Problem is Solved
To find the probability that at least one student solves the problem, we subtract the probability that none solve from 1:
- Probability(At least one solves) = 1 - Probability(None solve)
- Probability(At least one solves) = 1 - 1/4 = 3/4
Conclusion
Thus, the probability that the problem will be solved by at least one of the students A, B, or C is:
- 3/4
This matches the correct answer option 'D'.
A shelf has 6 mathematics books and 4 physics books. The probability that 3 particular mathematics books will be together is?- a)1/15
- b)2/15
- c)3/15
- d)4/15
Correct answer is option 'A'. Can you explain this answer?
A shelf has 6 mathematics books and 4 physics books. The probability that 3 particular mathematics books will be together is?
a)
1/15
b)
2/15
c)
3/15
d)
4/15
|
|
Chirag Verma answered |
The three particular books are treated as a unit which are arranged among the m selves after total 4 + 3 + 1 = 8 units have been arranged.
You have three coins in your pocket, two fair ones but the third biased with probability of heads p and tails 1 - p. One coin selected at random drops to the floor, landing heads up. How likely is it that it is one of the fair coins?- a)p
- b)1/(1+p)
- c)1/2
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
You have three coins in your pocket, two fair ones but the third biased with probability of heads p and tails 1 - p. One coin selected at random drops to the floor, landing heads up. How likely is it that it is one of the fair coins?
a)
p
b)
1/(1+p)
c)
1/2
d)
None of these
|
|
Chirag Verma answered |
One coin is selected at random and is thrown with head landing up.
By Bayes theorem for conditional probability
A : event that fair coin was selected
B : head has occurred
then
By Bayes theorem for conditional probability
A : event that fair coin was selected
B : head has occurred
then
A player is going to play a match either in the morning or in the afternoon or in the evening all possibilities being equally likely. The probability that he wins the match is 0.6, 0.1 and 0.8 according as if the match is played in the morning, afternoon or in the evening respectively. Given that he has won the match, the probability that the match was played in the afternoon is- a)1/12
- b)1/15
- c)2/27
- d)1/10
- e)1/20
Correct answer is option 'B'. Can you explain this answer?
A player is going to play a match either in the morning or in the afternoon or in the evening all possibilities being equally likely. The probability that he wins the match is 0.6, 0.1 and 0.8 according as if the match is played in the morning, afternoon or in the evening respectively. Given that he has won the match, the probability that the match was played in the afternoon is
a)
1/12
b)
1/15
c)
2/27
d)
1/10
e)
1/20
|
|
Chirag Verma answered |
This is the problem of conditional probability, so by Bave’s theorem.
A = the game was played in afternoonB B = player won the game.
A = the game was played in afternoonB B = player won the game.
Two dice are thrown once the probability of getting a sum 9 is given by :- a)1/12
- b)1/18
- c)1/6
- d)N.O.T.
Correct answer is option 'D'. Can you explain this answer?
Two dice are thrown once the probability of getting a sum 9 is given by :
a)
1/12
b)
1/18
c)
1/6
d)
N.O.T.
|
|
Veda Institute answered |
For sum of 9, the following values of the number on the faces are possible (3, 6) ; (4, 5) ; (5, 4) and (6, 3)
So, probability of getting sum 9 is
So, probability of getting sum 9 is
For a binomial distribution n = 40 and P = 0.05. The variance for the data is :- a)2.4
- b)3.2
- c)1.24
- d)1.9
Correct answer is option 'D'. Can you explain this answer?
For a binomial distribution n = 40 and P = 0.05. The variance for the data is :
a)
2.4
b)
3.2
c)
1.24
d)
1.9
|
|
Naveen Patel answered |
We know that ncr p(power x) and q(power n-x)
here p= no of success and
q= no of failures
n= no.of event
p+q=1
we want varience then E(x)=x * binomial varite
E(x2)=x2*binomial varient
we want to find varience (x) = (E(x2)-E(x)E(x))
after simplification we will get varience is equal to= 1.9
Let X be the random variable giving the number of heads obtained in 162 successive tosses of a biased coin with probability of getting head in a toss is 1/3. Assume that the losses are in dependent. The standard deviation of X is.- a)6
- b)8
- c)7
- d)9
Correct answer is option 'A'. Can you explain this answer?
Let X be the random variable giving the number of heads obtained in 162 successive tosses of a biased coin with probability of getting head in a toss is 1/3. Assume that the losses are in dependent. The standard deviation of X is.
a)
6
b)
8
c)
7
d)
9
|
|
Devansh Choudhary answered |
Given:
- Random variable X represents the number of heads obtained in 162 successive tosses of a biased coin, where the probability of getting a head in a toss is 1/3.
- The tosses are assumed to be independent.
To find:
The standard deviation of X.
Solution:
Step 1: Determine the distribution of X
Since X represents the number of heads obtained in 162 tosses of a biased coin, X follows a binomial distribution with parameters n = 162 (number of trials) and p = 1/3 (probability of success - getting a head).
Step 2: Calculate the mean of X
The mean of a binomial distribution is given by the formula:
mean (μ) = n * p
Substituting the given values: μ = 162 * (1/3) = 54
Step 3: Calculate the variance of X
The variance of a binomial distribution is given by the formula:
variance (σ^2) = n * p * (1 - p)
Substituting the given values: σ^2 = 162 * (1/3) * (1 - 1/3) = 108 * (2/3) = 72
Step 4: Calculate the standard deviation of X
The standard deviation (σ) is the square root of the variance.
Taking the square root of the variance calculated in the previous step: σ = √72 = 6√2
Step 5: Simplify the standard deviation
To simplify the standard deviation, approximate the value of √2 to a decimal approximation:
√2 ≈ 1.414
Therefore, the standard deviation of X is approximately equal to 6 * 1.414 = 8.484
Answer: The correct option is A) 6.
- Random variable X represents the number of heads obtained in 162 successive tosses of a biased coin, where the probability of getting a head in a toss is 1/3.
- The tosses are assumed to be independent.
To find:
The standard deviation of X.
Solution:
Step 1: Determine the distribution of X
Since X represents the number of heads obtained in 162 tosses of a biased coin, X follows a binomial distribution with parameters n = 162 (number of trials) and p = 1/3 (probability of success - getting a head).
Step 2: Calculate the mean of X
The mean of a binomial distribution is given by the formula:
mean (μ) = n * p
Substituting the given values: μ = 162 * (1/3) = 54
Step 3: Calculate the variance of X
The variance of a binomial distribution is given by the formula:
variance (σ^2) = n * p * (1 - p)
Substituting the given values: σ^2 = 162 * (1/3) * (1 - 1/3) = 108 * (2/3) = 72
Step 4: Calculate the standard deviation of X
The standard deviation (σ) is the square root of the variance.
Taking the square root of the variance calculated in the previous step: σ = √72 = 6√2
Step 5: Simplify the standard deviation
To simplify the standard deviation, approximate the value of √2 to a decimal approximation:
√2 ≈ 1.414
Therefore, the standard deviation of X is approximately equal to 6 * 1.414 = 8.484
Answer: The correct option is A) 6.
If A ∩ B = φ and B ∩ C = φ, then P(A∪B∪C) = - a)P(A) + P(B) + P(C)
- b) P(A) P(B) P(C)
- c)P(A)P(B) + P(B)P(C) + P(C)P(A)
- d)P(A∩B) + P(B∩C)
Correct answer is option 'A'. Can you explain this answer?
If A ∩ B = φ and B ∩ C = φ, then P(A∪B∪C) =
a)
P(A) + P(B) + P(C)
b)
P(A) P(B) P(C)
c)
P(A)P(B) + P(B)P(C) + P(C)P(A)
d)
P(A∩B) + P(B∩C)
|
Tanvi Sengupta answered |
Explanation:
Given:
A∩B = ∅ (empty set)
B∩C = ∅ (empty set)
To find:
P(A∪B∪C)
Explanation:
When A∩B = ∅ and B∩C = ∅, it means that the events A, B, and C are mutually exclusive. This implies that the events do not have any outcomes in common.
When events are mutually exclusive, the probability of the union of these events is the sum of the individual probabilities of the events. This is because the probability of the union of mutually exclusive events is the sum of the probabilities of the individual events.
Therefore, P(A∪B∪C) = P(A) + P(B) + P(C)
Hence, the correct answer is option 'A': P(A) + P(B) + P(C)
Given:
A∩B = ∅ (empty set)
B∩C = ∅ (empty set)
To find:
P(A∪B∪C)
Explanation:
When A∩B = ∅ and B∩C = ∅, it means that the events A, B, and C are mutually exclusive. This implies that the events do not have any outcomes in common.
When events are mutually exclusive, the probability of the union of these events is the sum of the individual probabilities of the events. This is because the probability of the union of mutually exclusive events is the sum of the probabilities of the individual events.
Therefore, P(A∪B∪C) = P(A) + P(B) + P(C)
Hence, the correct answer is option 'A': P(A) + P(B) + P(C)
A speaks truth 3 times out of 4 and B speaks 7 times out of 10. They both assert that a white ball has been drawn from a bag containing 6 different colour balls. Find the probability of the truth of the assertion- a)21/40
- b)35/36
- c)39/40
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
A speaks truth 3 times out of 4 and B speaks 7 times out of 10. They both assert that a white ball has been drawn from a bag containing 6 different colour balls. Find the probability of the truth of the assertion
a)
21/40
b)
35/36
c)
39/40
d)
None of these
|
|
Chirag Verma answered |
If A has asserted that white ball is drawn. If white ball is drawn then he speaks truth else, he is lying.
Similar is the case with B.
i.e. (P(W, T, T) / P[W,T, T,) + P(N, F. F))
A letter is knwon to have come from either TATANAGAR' or CALCUTTA. On the envelope, Just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is.- a)4/11
- b)1/3
- c)5/12
- d)none
Correct answer is option 'A'. Can you explain this answer?
A letter is knwon to have come from either TATANAGAR' or CALCUTTA. On the envelope, Just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is.
a)
4/11
b)
1/3
c)
5/12
d)
none
|
|
Chirag Verma answered |
TATANAGAR has 9 letters, so number of ways in which two consecutive letters can be printed will be, 8 out of which there are 2 ways in which TA can be printed.
Similarly for CALCUTTA, there are 7 ways of printing two consecutive letters, from which there is only way to print TA'
So, required probability’
Similarly for CALCUTTA, there are 7 ways of printing two consecutive letters, from which there is only way to print TA'
So, required probability’
Two letters are chosen one after another without replacement from the English alphabet. What is the probability that the second letter chosen is a vowel?- a)4/25
- b)5/26
- c)5/25
- d)1/5,1/26
Correct answer is option 'B'. Can you explain this answer?
Two letters are chosen one after another without replacement from the English alphabet. What is the probability that the second letter chosen is a vowel?
a)
4/25
b)
5/26
c)
5/25
d)
1/5,1/26
|
|
Veda Institute answered |
Note : There are 21 cosonants and 5 vowels in 26 letters of alphabet.
Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:- a)1 (1-pr)n-1
- b)(1-pr)n-1
- c)(pr)n-1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:
a)
1 (1-pr)n-1
b)
(1-pr)n-1
c)
(pr)n-1
d)
None of these
|
|
Chirag Verma answered |
At least one missile will remain in reserve implies that upto (n - 1)th hit target has been brought down.
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
A and B throw with one die for a stake of Rs 11 which is to be won by the player who first throws 6. If A has the first throw, what is the expectation of his winning the stake?- a)5/11
- b)15/26
- c)6/11
- d)16/26
Correct answer is option 'C'. Can you explain this answer?
A and B throw with one die for a stake of Rs 11 which is to be won by the player who first throws 6. If A has the first throw, what is the expectation of his winning the stake?
a)
5/11
b)
15/26
c)
6/11
d)
16/26
|
|
Chirag Verma answered |
Probability of obtaining 6 in a single throw is 1/6. Probability that A wins the stake if he get 6 in first throw or if he loses then B also lose and then he get 6 in his second throw i.e. overal 3rd throw and so on.
[sum of infinite terms of G.P.]
[sum of infinite terms of G.P.]At a railway station a passenger leaves his luggage in a locker which is opened by dialling a three-digit code. The passenger chooses the code, closes the locker and leaves for the town. A stranger who does not know the code tries to open the locker by dialling three digits at random without repeating the unsuccessful codes. The probability that the stranger does not have to perform more than 100 trials for a success is- a)9/10
- b)999/1000
- c)1/1000
- d)1/10
Correct answer is option 'D'. Can you explain this answer?
At a railway station a passenger leaves his luggage in a locker which is opened by dialling a three-digit code. The passenger chooses the code, closes the locker and leaves for the town. A stranger who does not know the code tries to open the locker by dialling three digits at random without repeating the unsuccessful codes. The probability that the stranger does not have to perform more than 100 trials for a success is
a)
9/10
b)
999/1000
c)
1/1000
d)
1/10
|
|
Chirag Verma answered |
j = 0, 1, 2,...
Find the value of k in the equation x3 -6x2 + kx + 64 = 0, if it is known that the roots of the equation are in geometric progression.- a)24
- b)16
- c)-16
- d)-24
Correct answer is option 'D'. Can you explain this answer?
Find the value of k in the equation x3 -6x2 + kx + 64 = 0, if it is known that the roots of the equation are in geometric progression.
a)
24
b)
16
c)
-16
d)
-24
|
|
Chirag Verma answered |
Let a/r, a and ar be the three roots, then
Now, sum of the roots taken two at a time
2(-4) + (-4) (8) + 8(2) = K
⇒ K = -24
Now, sum of the roots taken two at a time
2(-4) + (-4) (8) + 8(2) = K
⇒ K = -24
Prob. of getting an odd number or a no. less than 4 in throwing a dice is :- a)1/3
- b)2/3
- c)1/2
- d)3/5
Correct answer is option 'B'. Can you explain this answer?
Prob. of getting an odd number or a no. less than 4 in throwing a dice is :
a)
1/3
b)
2/3
c)
1/2
d)
3/5
|
|
Veda Institute answered |
The number which is odd or less than 4 is in our favour. So out of 6 numbers on the dice 4 numbers are in our favour (i.e. 1, 2, 3, 5) So probability 
If two dice, are tossed the probability of getting the sum at least 5 is- a)7/12
- b)11/12
- c)1/2
- d)5/6
Correct answer is option 'D'. Can you explain this answer?
If two dice, are tossed the probability of getting the sum at least 5 is
a)
7/12
b)
11/12
c)
1/2
d)
5/6
|
|
Veda Institute answered |
Sum of the number so n the face so f two dice, is 2, 3 and 4 in 1 , 2 and 3 cases respectively, so remaining 36-(1+2+3) - 30 cases are in our favour.
So, probability =
So, probability =
What is the probability of getting seventh head in the tenth toss of an unbiased coin.- a)21/256
- b)15/128
- c)21/128
- d)15/256
Correct answer is option 'A'. Can you explain this answer?
What is the probability of getting seventh head in the tenth toss of an unbiased coin.
a)
21/256
b)
15/128
c)
21/128
d)
15/256
|
Anika Gupta answered |
To find the probability of getting the seventh head in the tenth toss of an unbiased coin, we can use the concept of binomial distribution.
**Binomial Distribution:**
A binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials. In this case, the coin tosses can be considered as Bernoulli trials, where the outcome of each trial is either a head or a tail.
**Probability of getting a head:**
Since the coin is unbiased, the probability of getting a head (H) is 0.5, and the probability of getting a tail (T) is also 0.5.
**Calculating the probability:**
To calculate the probability of getting the seventh head in the tenth toss, we need to consider the different possible arrangements of heads and tails in the first nine tosses.
**Case 1:**
If we get exactly six heads in the first nine tosses, the seventh toss must be a head. The probability of getting six heads in nine tosses can be calculated using the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
P(X = k) is the probability of getting k successes,
n is the number of trials,
k is the number of successes,
p is the probability of success.
In this case, n = 9, k = 6, and p = 0.5. Plugging these values into the formula:
P(X = 6) = C(9, 6) * (0.5)^6 * (1-0.5)^(9-6)
= 84 * 0.015625 * 0.125
= 0.13125
The probability of getting exactly six heads in the first nine tosses is 0.13125.
**Case 2:**
If we get exactly seven heads in the first nine tosses, the tenth toss must also be a head. The probability of getting seven heads in nine tosses can be calculated using the same binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
In this case, n = 9, k = 7, and p = 0.5. Plugging these values into the formula:
P(X = 7) = C(9, 7) * (0.5)^7 * (1-0.5)^(9-7)
= 36 * 0.0078125 * 0.25
= 0.0703125
The probability of getting exactly seven heads in the first nine tosses is 0.0703125.
**Case 3:**
If we get eight or more heads in the first nine tosses, the tenth toss will not be the seventh head. Therefore, the probability in this case is 0.
**Total probability:**
To find the total probability of getting the seventh head in the tenth toss, we need to add the probabilities from Case 1 and Case 2:
P(seventh head in tenth toss) = P(X = 6) + P(X = 7)
= 0.13125 + 0.0703125
= 0.2015625
Converting this
**Binomial Distribution:**
A binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials. In this case, the coin tosses can be considered as Bernoulli trials, where the outcome of each trial is either a head or a tail.
**Probability of getting a head:**
Since the coin is unbiased, the probability of getting a head (H) is 0.5, and the probability of getting a tail (T) is also 0.5.
**Calculating the probability:**
To calculate the probability of getting the seventh head in the tenth toss, we need to consider the different possible arrangements of heads and tails in the first nine tosses.
**Case 1:**
If we get exactly six heads in the first nine tosses, the seventh toss must be a head. The probability of getting six heads in nine tosses can be calculated using the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
P(X = k) is the probability of getting k successes,
n is the number of trials,
k is the number of successes,
p is the probability of success.
In this case, n = 9, k = 6, and p = 0.5. Plugging these values into the formula:
P(X = 6) = C(9, 6) * (0.5)^6 * (1-0.5)^(9-6)
= 84 * 0.015625 * 0.125
= 0.13125
The probability of getting exactly six heads in the first nine tosses is 0.13125.
**Case 2:**
If we get exactly seven heads in the first nine tosses, the tenth toss must also be a head. The probability of getting seven heads in nine tosses can be calculated using the same binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
In this case, n = 9, k = 7, and p = 0.5. Plugging these values into the formula:
P(X = 7) = C(9, 7) * (0.5)^7 * (1-0.5)^(9-7)
= 36 * 0.0078125 * 0.25
= 0.0703125
The probability of getting exactly seven heads in the first nine tosses is 0.0703125.
**Case 3:**
If we get eight or more heads in the first nine tosses, the tenth toss will not be the seventh head. Therefore, the probability in this case is 0.
**Total probability:**
To find the total probability of getting the seventh head in the tenth toss, we need to add the probabilities from Case 1 and Case 2:
P(seventh head in tenth toss) = P(X = 6) + P(X = 7)
= 0.13125 + 0.0703125
= 0.2015625
Converting this
A six faced die is a biased one. it is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even, is.- a)4/8
- b)5/8
- c)6/8
- d)7/8
Correct answer is option 'B'. Can you explain this answer?
A six faced die is a biased one. it is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even, is.
a)
4/8
b)
5/8
c)
6/8
d)
7/8
|
|
Chirag Verma answered |
Now sum of two odd numbers, and also sum of two even numbers is even, so.
P (odd, odd) + P (even, even)
The probability that a radar will detect an object in one cycle is p. The probability that it will be detected in n cycles is- a)1-pn
- b)1-(1-p)n
- c)pn
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The probability that a radar will detect an object in one cycle is p. The probability that it will be detected in n cycles is
a)
1-pn
b)
1-(1-p)n
c)
pn
d)
None of these
|
|
Chirag Verma answered |
1 - prob. than in n cycle sit is not detected,
= 1-(1-p)n
= 1-(1-p)n
There are two groups of subjects, one of which consists of 5 science and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects, an unbiased die is cast. If the number 3 or 5 turns up a subject is selected at random from the first group. What is the probability that an engineering subject is ultimately selected?- a)13/24
- b)5/12
- c)9/16
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
There are two groups of subjects, one of which consists of 5 science and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects, an unbiased die is cast. If the number 3 or 5 turns up a subject is selected at random from the first group. What is the probability that an engineering subject is ultimately selected?
a)
13/24
b)
5/12
c)
9/16
d)
None of these
|
|
Trisha Sharma answered |
Understanding the Problem
To find the probability of selecting an engineering subject after rolling an unbiased die, we need to consider the two groups of subjects and the outcomes of the die roll.
Die Outcomes
- The die has 6 faces with numbers 1 to 6.
- Outcomes that lead to selecting from Group 1: 3 or 5 (2 favorable outcomes).
- Probability of rolling a 3 or 5:
- P(3 or 5) = Number of favorable outcomes / Total outcomes = 2/6 = 1/3
Group 1 Composition
- Group 1 consists of 5 science and 3 engineering subjects.
- Total subjects in Group 1 = 5 + 3 = 8.
- Probability of selecting an engineering subject from Group 1:
- P(Engineering | Group 1) = Number of engineering subjects / Total subjects = 3/8
Calculating Overall Probability
To find the overall probability of selecting an engineering subject when the die shows 3 or 5 and we choose from Group 1, we multiply the probabilities:
- Overall Probability = P(3 or 5) * P(Engineering | Group 1)
- Overall Probability = (1/3) * (3/8) = 3/24 = 1/8
Conclusion
However, we also need to consider the probability of not rolling a 3 or 5, which would lead to selecting from Group 2. The outcomes leading to Group 2 are 1, 2, 4, or 6 (4 favorable outcomes).
Group 2 Composition
- Group 2 consists of 3 science and 5 engineering subjects.
- Total subjects in Group 2 = 3 + 5 = 8.
- Probability of selecting an engineering subject from Group 2:
- P(Engineering | Group 2) = 5/8
Final Calculation
- Probability of not rolling a 3 or 5: P(Not 3 or 5) = 4/6 = 2/3
- Overall Probability from Group 2 = (2/3) * (5/8) = 10/24
Total Probability of Selecting Engineering Subject
- Total Probability = Probability from Group 1 + Probability from Group 2
- Total Probability = (3/24) + (10/24) = 13/24
Thus, the correct answer is option 'A': 13/24.
To find the probability of selecting an engineering subject after rolling an unbiased die, we need to consider the two groups of subjects and the outcomes of the die roll.
Die Outcomes
- The die has 6 faces with numbers 1 to 6.
- Outcomes that lead to selecting from Group 1: 3 or 5 (2 favorable outcomes).
- Probability of rolling a 3 or 5:
- P(3 or 5) = Number of favorable outcomes / Total outcomes = 2/6 = 1/3
Group 1 Composition
- Group 1 consists of 5 science and 3 engineering subjects.
- Total subjects in Group 1 = 5 + 3 = 8.
- Probability of selecting an engineering subject from Group 1:
- P(Engineering | Group 1) = Number of engineering subjects / Total subjects = 3/8
Calculating Overall Probability
To find the overall probability of selecting an engineering subject when the die shows 3 or 5 and we choose from Group 1, we multiply the probabilities:
- Overall Probability = P(3 or 5) * P(Engineering | Group 1)
- Overall Probability = (1/3) * (3/8) = 3/24 = 1/8
Conclusion
However, we also need to consider the probability of not rolling a 3 or 5, which would lead to selecting from Group 2. The outcomes leading to Group 2 are 1, 2, 4, or 6 (4 favorable outcomes).
Group 2 Composition
- Group 2 consists of 3 science and 5 engineering subjects.
- Total subjects in Group 2 = 3 + 5 = 8.
- Probability of selecting an engineering subject from Group 2:
- P(Engineering | Group 2) = 5/8
Final Calculation
- Probability of not rolling a 3 or 5: P(Not 3 or 5) = 4/6 = 2/3
- Overall Probability from Group 2 = (2/3) * (5/8) = 10/24
Total Probability of Selecting Engineering Subject
- Total Probability = Probability from Group 1 + Probability from Group 2
- Total Probability = (3/24) + (10/24) = 13/24
Thus, the correct answer is option 'A': 13/24.
Suppose there are 15 different types of coupons and suppose that each time one obtains a coupon, it is equally likely to be any one of the 15 types. The expected number of different types that are contained in the set of 5 coupons is- a)
- b)
- c)
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Suppose there are 15 different types of coupons and suppose that each time one obtains a coupon, it is equally likely to be any one of the 15 types. The expected number of different types that are contained in the set of 5 coupons is
a)
b)
c)
d)
none of these
|
|
Chirag Verma answered |
Probability that any particular coupon will be selected is 1/15.
Expected number of different types of coupons =
Expected number of different types of coupons =
If the letters of the word REGULATION' be arranged at random, the probability that there will be exactly 4 letters between R and E is- a)1/10
- b)1/9
- c)1/5
- d)1/2
Correct answer is option 'B'. Can you explain this answer?
If the letters of the word REGULATION' be arranged at random, the probability that there will be exactly 4 letters between R and E is
a)
1/10
b)
1/9
c)
1/5
d)
1/2
|
|
Chirag Verma answered |
R E G U L A T IO N
It has 10 letters. Four letters are there between R and E, if there positions are
1, 6; 2, 7; 3, 8; 4, 9 or 5, 10 i.e. 5 ways
So, required probability is
It has 10 letters. Four letters are there between R and E, if there positions are
1, 6; 2, 7; 3, 8; 4, 9 or 5, 10 i.e. 5 ways
So, required probability is
If a student is likely to choose any of the four choices with equal probability in a multiple-choice examination with five questions then the probability that the student answer at least four questions correctly is- a)16/45
- b)15/45
- c)4/45
- d)1/45
Correct answer is option 'A'. Can you explain this answer?
If a student is likely to choose any of the four choices with equal probability in a multiple-choice examination with five questions then the probability that the student answer at least four questions correctly is
a)
16/45
b)
15/45
c)
4/45
d)
1/45
|
|
Chirag Verma answered |
P(Answering at least 4 questions correctly)
Chapter doubts & questions for Probability and Statistics - Topic-wise Tests & Solved Examples for Mathematics 2025 is part of Mathematics exam preparation. The chapters have been prepared according to the Mathematics exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mathematics 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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