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What is the probability that a divisor of 1099 is a multiple of 1096?- a)1/625
- b)4/625
- c)12/625
- d)16/625
Correct answer is option 'A'. Can you explain this answer?
What is the probability that a divisor of 1099 is a multiple of 1096?
a)
1/625
b)
4/625
c)
12/625
d)
16/625
|
Chirag Verma answered |
p(multiple of 10% |divisor of 1099)
Since 10 = 2.5
1099 = 299 . 599
Any divisor of 1099 is of the form 2a . 5b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99.
The number of such possibilities is combination of 100 values of a and 100 values of b = 100 × 100 each of which is a divisor of 1099.
So, no. of divisors of 1099 = 100 × 100.
Any number which is a multiple of 1096 as well as divisor of 1099 is of the form 2a . 5b where 96 ≤ a ≤ 99 and 96 ≤ b ≤ 99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 1096 as well as a divisor of 1099.
∴ p(multiple of 1096|divisor of 1099)
Since 10 = 2.5
1099 = 299 . 599
Any divisor of 1099 is of the form 2a . 5b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99.
The number of such possibilities is combination of 100 values of a and 100 values of b = 100 × 100 each of which is a divisor of 1099.
So, no. of divisors of 1099 = 100 × 100.
Any number which is a multiple of 1096 as well as divisor of 1099 is of the form 2a . 5b where 96 ≤ a ≤ 99 and 96 ≤ b ≤ 99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 1096 as well as a divisor of 1099.
∴ p(multiple of 1096|divisor of 1099)
You have gone to a cyber-café with a friend. You found that the cyber-cafe has only three terminals. All terminals are unoccupied. You and your friend have to make a random choice of selecting a terminal. What is the probability that both of you will NOT select the same terminal?- a)1/9
- b)1/3
- c)2/3
- d)1
Correct answer is option 'C'. Can you explain this answer?
You have gone to a cyber-café with a friend. You found that the cyber-cafe has only three terminals. All terminals are unoccupied. You and your friend have to make a random choice of selecting a terminal. What is the probability that both of you will NOT select the same terminal?
a)
1/9
b)
1/3
c)
2/3
d)
1
|
|
Chirag Verma answered |
Out of three terminals probability of selecting terminals of two friends is = 1/3
∴ Probability of not selecting same terminal = 1- 1/3 = 2/3
∴ Probability of not selecting same terminal = 1- 1/3 = 2/3
Let X be a randon variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean –1 and variance unknown. If P(X ≤ –1) = P(Y ≥ 2) the standard deviation of Y is- a)3
- b)2
- c)√2
- d)1
Correct answer is option 'A'. Can you explain this answer?
Let X be a randon variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean –1 and variance unknown. If P(X ≤ –1) = P(Y ≥ 2) the standard deviation of Y is
a)
3
b)
2
c)
√2
d)
1
|
|
Chirag Verma answered |
given
Convert into standard and normal variates,
...(i)Now since us know that in standard normal distribution,
Aishwarya studies either computer science or mathematics everyday. if the studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?- a)0.24
- b)0.36
- c)0.4
- d)0.6
Correct answer is option 'C'. Can you explain this answer?
Aishwarya studies either computer science or mathematics everyday. if the studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
a)
0.24
b)
0.36
c)
0.4
d)
0.6
|
|
Chirag Verma answered |
Let C denote computes science study and M denotes maths study.
P(C on monday and C on wednesday) + p(C on monday, c on tuesday and C on wednesday)
= 1 x 0.6 x 0.4 + 1 x 0.4 x 0.4
= 0.24 + 0.16
= 0.40
P(C on monday and C on wednesday) + p(C on monday, c on tuesday and C on wednesday)
= 1 x 0.6 x 0.4 + 1 x 0.4 x 0.4
= 0.24 + 0.16
= 0.40
Poisson's ratio for a metal is 0.35. Neglecting piezo-resistance effect, the gage factor of a strain gage made of this metal is- a)0.65
- b)1
- c)1.35
- d)1.70
Correct answer is option 'D'. Can you explain this answer?
Poisson's ratio for a metal is 0.35. Neglecting piezo-resistance effect, the gage factor of a strain gage made of this metal is
a)
0.65
b)
1
c)
1.35
d)
1.70
|
Aryan Verma answered |
Explanation:
Poisson's ratio:
- Poisson's ratio (ν) is a measure of the amount of transverse strain that results from a given axial strain.
- It is given by the formula: ν = -ε_transverse / ε_axial, where ε is strain.
Gage factor:
- The gage factor (GF) of a strain gage is defined as the ratio of the relative change in resistance (ΔR/R) to the strain (ε) being measured.
- It is given by the formula: GF = ΔR / R / ε.
Calculation:
- Poisson's ratio (ν) = 0.35
- For a metal, the relationship between Poisson's ratio and gage factor is given by the formula: GF = 2(1 + ν)
- Substituting the value of Poisson's ratio into the formula, we get: GF = 2(1 + 0.35) = 2(1.35) = 2.70
Therefore, the gage factor of a strain gage made of this metal is 1.70. The correct answer is option 'D'.
Poisson's ratio:
- Poisson's ratio (ν) is a measure of the amount of transverse strain that results from a given axial strain.
- It is given by the formula: ν = -ε_transverse / ε_axial, where ε is strain.
Gage factor:
- The gage factor (GF) of a strain gage is defined as the ratio of the relative change in resistance (ΔR/R) to the strain (ε) being measured.
- It is given by the formula: GF = ΔR / R / ε.
Calculation:
- Poisson's ratio (ν) = 0.35
- For a metal, the relationship between Poisson's ratio and gage factor is given by the formula: GF = 2(1 + ν)
- Substituting the value of Poisson's ratio into the formula, we get: GF = 2(1 + 0.35) = 2(1.35) = 2.70
Therefore, the gage factor of a strain gage made of this metal is 1.70. The correct answer is option 'D'.
Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3 ….., 20. What is the probability that 2 appears at an earlier position that any other even number in the selected permutation? - a)1/2
- b)1/10
- c)9!/20!
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3 ….., 20. What is the probability that 2 appears at an earlier position that any other even number in the selected permutation?
a)
1/2
b)
1/10
c)
9!/20!
d)
None of these
|
Veda Institute answered |
Number of permutations with ‘2’ in the first position = 19!
Number of permutations with ‘2’ in the second position = 10 × 18!
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
Number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by
Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.
Number of permutations with ‘2’ in the second position = 10 × 18!
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
Number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by
Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.
An unbalanced ice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same.
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closed to the probability that the face value exceeds 3?- a)0.453
- b)0.468
- c)0.485
- d)0.492
Correct answer is option 'B'. Can you explain this answer?
An unbalanced ice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same.
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closed to the probability that the face value exceeds 3?
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closed to the probability that the face value exceeds 3?
a)
0.453
b)
0.468
c)
0.485
d)
0.492
|
|
Chirag Verma answered |
It is given that
P(odd) = 0.9 p(even)
Now since ∑p(x) = 1
∴ p(odd) + p(even) = 1
⇒ 0.9p(even) + p(even) = 1
⇒ p(even) = 1/1.9 = 0.5263
Now, it is given that p (any even face) is same
i.e p(2) = p(4) = p(6)
Now since,
∴
= 1/3 (0.5263)
= 0.1754
It is given that
P(odd) = 0.9 p(even)
Now since ∑p(x) = 1
∴ p(odd) + p(even) = 1
⇒ 0.9p(even) + p(even) = 1
⇒ p(even) = 1/1.9 = 0.5263
Now, it is given that p (any even face) is same
i.e p(2) = p(4) = p(6)
Now since,
∴
= 1/3 (0.5263)
= 0.1754
It is given that
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company, therefore, subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q.
What is the probability of a computer being declared faulty?- a)pq + (1-p)(1-q)
- b)(1-q)p
- c)(1-p)q
- d)pq
Correct answer is option 'A'. Can you explain this answer?
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company, therefore, subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q.
What is the probability of a computer being declared faulty?
What is the probability of a computer being declared faulty?
a)
pq + (1-p)(1-q)
b)
(1-q)p
c)
(1-p)q
d)
pq
|
Devansh Choudhary answered |
Probability of a computer being declared faulty
To calculate the probability of a computer being declared faulty, we need to consider the probabilities of two independent events: the assembly of a faulty computer and the testing process giving the correct result.
Given Information:
- Probability of a faulty assembly of any computer: p
- Probability of the testing process giving the correct result: q
Approach:
To find the probability of a computer being declared faulty, we need to consider two scenarios:
1. The computer is faulty and the testing process gives the correct result.
2. The computer is not faulty and the testing process gives an incorrect result.
Scenario 1: The computer is faulty and the testing process gives the correct result.
The probability of a computer being faulty is p, and the probability of the testing process giving the correct result is q. Since these two events are independent, we can multiply their probabilities to find the probability of both events occurring simultaneously. Therefore, the probability of a computer being faulty and the testing process giving the correct result is pq.
Scenario 2: The computer is not faulty and the testing process gives an incorrect result.
The probability of a computer not being faulty is (1 - p), and the probability of the testing process giving an incorrect result is (1 - q). Again, since these two events are independent, we can multiply their probabilities to find the probability of both events occurring simultaneously. Therefore, the probability of a computer not being faulty and the testing process giving an incorrect result is (1 - p)(1 - q).
Conclusion:
The probability of a computer being declared faulty is the sum of the probabilities from both scenarios:
- Scenario 1: pq
- Scenario 2: (1 - p)(1 - q)
Therefore, the total probability of a computer being declared faulty is pq + (1 - p)(1 - q).
To calculate the probability of a computer being declared faulty, we need to consider the probabilities of two independent events: the assembly of a faulty computer and the testing process giving the correct result.
Given Information:
- Probability of a faulty assembly of any computer: p
- Probability of the testing process giving the correct result: q
Approach:
To find the probability of a computer being declared faulty, we need to consider two scenarios:
1. The computer is faulty and the testing process gives the correct result.
2. The computer is not faulty and the testing process gives an incorrect result.
Scenario 1: The computer is faulty and the testing process gives the correct result.
The probability of a computer being faulty is p, and the probability of the testing process giving the correct result is q. Since these two events are independent, we can multiply their probabilities to find the probability of both events occurring simultaneously. Therefore, the probability of a computer being faulty and the testing process giving the correct result is pq.
Scenario 2: The computer is not faulty and the testing process gives an incorrect result.
The probability of a computer not being faulty is (1 - p), and the probability of the testing process giving an incorrect result is (1 - q). Again, since these two events are independent, we can multiply their probabilities to find the probability of both events occurring simultaneously. Therefore, the probability of a computer not being faulty and the testing process giving an incorrect result is (1 - p)(1 - q).
Conclusion:
The probability of a computer being declared faulty is the sum of the probabilities from both scenarios:
- Scenario 1: pq
- Scenario 2: (1 - p)(1 - q)
Therefore, the total probability of a computer being declared faulty is pq + (1 - p)(1 - q).
If the difference between the expectation of the square of a random variable ( E [X])2 is denoted by R, then- a)R = 0
- b)R< 0
- c)R ≥ 0
- d)R > 0
Correct answer is option 'C'. Can you explain this answer?
If the difference between the expectation of the square of a random variable ( E [X])2 is denoted by R, then
a)
R = 0
b)
R< 0
c)
R ≥ 0
d)
R > 0
|
Aditi Singh answered |
= 0 implies that the variable X has no variance or variability. In other words, it means that X always takes on the same value, so there is no difference between the expected value of X squared and the square of the expected value of X. This is a special case and not true for most random variables.
b) R > 0 implies that the variable X has some variance or variability. In this case, there is a difference between the expected value of X squared and the square of the expected value of X. This is the more typical scenario for random variables.
b) R > 0 implies that the variable X has some variance or variability. In this case, there is a difference between the expected value of X squared and the square of the expected value of X. This is the more typical scenario for random variables.
An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches – 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained all these students is- a)0
- b)2550
- c)7525
- d)9375
Correct answer is option 'D'. Can you explain this answer?
An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches – 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained all these students is
a)
0
b)
2550
c)
7525
d)
9375
|
|
Chirag Verma answered |
Let the marks obtained per question be a random variable X. It’s probability distribution table is given below:
For each element is a set of size 2n, an unbiased coin is tossed. The 2n coin tossed are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is
- a)
- b)
- c)
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
For each element is a set of size 2n, an unbiased coin is tossed. The 2n coin tossed are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is
a)
b)
c)
d)
1/2
|
|
Chirag Verma answered |
The probability that exactly n elements are chosen = the probability of getting n heads out of 2n tosses
The probability that the conversation will exceed five minutes is
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?- a)3/8
- b)1/2
- c)5/8
- d)3/4
Correct answer is option 'A'. Can you explain this answer?
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?
a)
3/8
b)
1/2
c)
5/8
d)
3/4
|
|
Veda Institute answered |
Here P(H) = P(T) = 1/2
It's a Bernoulli's trials.
∴ Required probability
It's a Bernoulli's trials.
∴ Required probability
A program consists of two modules executed sequentially. Let f1 (t) and f2 (t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by- a)f1( t ) + f2( t )
- b)
- c)
- d)max {f1(t), f2(t)}
Correct answer is option 'C'. Can you explain this answer?
A program consists of two modules executed sequentially. Let f1 (t) and f2 (t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by
a)
f1( t ) + f2( t )
b)
c)
d)
max {f1(t), f2(t)}
|
|
Veda Institute answered |
Let the time taken for first and second modules be represented by x and y and
total time = t.
and y and total time = t.
∴ t = x + y is a random variable
Now the joint density function
which is also called as convolution of f1 and f2, abbreviated as f1* f2.
total time = t.
and y and total time = t.
∴ t = x + y is a random variable
Now the joint density function
which is also called as convolution of f1 and f2, abbreviated as f1* f2.
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is - a)1/3
- b)3/7
- c)1/2
- d)4/7
Correct answer is option 'C'. Can you explain this answer?
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is
a)
1/3
b)
3/7
c)
1/2
d)
4/7
|
|
Chirag Verma answered |
After first ball is drawn white then sample space has 4 + 3 – 1 = 6 balls.
Probability of second ball is red without replacement
Probability of second ball is red without replacement
A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is- a)1/8
- b)1/2
- c)3/8
- d)3/4
Correct answer is option 'D'. Can you explain this answer?
A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is
a)
1/8
b)
1/2
c)
3/8
d)
3/4
|
|
Chirag Verma answered |
After first head in first toss, probability of tails in 2nd and 3rd toss = 1/2.1/2 = 1/4
∴ Probability of exactly two heads = 1 - 1/4 = 3/4
∴ Probability of exactly two heads = 1 - 1/4 = 3/4
Let P(E) denote the probability of the even E. Given P(A) = 1, P(B) = 1/2, the values of P(A/B) and p(B/A) respectively are - a)1/4, 1/2
- b)1/2, 1/4
- c)1/2 , 1
- d)1, 1/2
Correct answer is option 'D'. Can you explain this answer?
Let P(E) denote the probability of the even E. Given P(A) = 1, P(B) = 1/2, the values of P(A/B) and p(B/A) respectively are
a)
1/4, 1/2
b)
1/2, 1/4
c)
1/2 , 1
d)
1, 1/2
|
|
Veda Institute answered |
Here, P(A) = 1, P(B) = 1/2
Since A, B are independent events,
∴ P(AB) = P(A)P(B)
Since A, B are independent events,
∴ P(AB) = P(A)P(B)
Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even, is- a)0.5
- b)0.25
- c)0.167
- d)0.125
Correct answer is option 'A'. Can you explain this answer?
Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even, is
a)
0.5
b)
0.25
c)
0.167
d)
0.125
|
|
Chirag Verma answered |
Here sample space S = 6 x 6 = 36
Total no. of way in which sum of digits on the top surface of the two dice is even is 18.
∴ The require probability = 18/36 = 0.5.
Total no. of way in which sum of digits on the top surface of the two dice is even is 18.
∴ The require probability = 18/36 = 0.5.
Using the given data points tabulated below, a straight line passing through the origin is fitted using least squares method. The slope os the line is
- a)0.9
- b)1.0
- c)1.1
- d)1.5
Correct answer is option 'C'. Can you explain this answer?
Using the given data points tabulated below, a straight line passing through the origin is fitted using least squares method. The slope os the line is
a)
0.9
b)
1.0
c)
1.1
d)
1.5
|
|
Veda Institute answered |
Suppose the line being, y = mx
Since, it has been fit by least square method, therefore
∑ y = μ ∑ x, and ∑ x y = μ ∑ x2
∴ m = 1.1
Since, it has been fit by least square method, therefore
∑ y = μ ∑ x, and ∑ x y = μ ∑ x2
∴ m = 1.1
The standard normal probability function can be approximated as
Where xN = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 and 102 cm is - a)66.7%
- b)50.0%
- c)33.3%
- d)16.7%
Correct answer is option 'D'. Can you explain this answer?
The standard normal probability function can be approximated as
Where xN = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 and 102 cm is
Where xN = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 and 102 cm is
a)
66.7%
b)
50.0%
c)
33.3%
d)
16.7%
|
|
Veda Institute answered |
Here μ = 102 cm and σ = 27 cm
This area is shown below
The shaded area in above figure is given by F(0) - F(-0.44)
= 0.5 - 0.3345
= 0.1655 = 16.55%
This area is shown below
The shaded area in above figure is given by F(0) - F(-0.44)
= 0.5 - 0.3345
= 0.1655 = 16.55%
Probability density function p(x) of a random variable x is as shown below. The value of α is
- a)2/c
- b)1/c
- c)2/(b+c)
- d)1/(b+c)
Correct answer is option 'A'. Can you explain this answer?
Probability density function p(x) of a random variable x is as shown below. The value of α is
a)
2/c
b)
1/c
c)
2/(b+c)
d)
1/(b+c)
|
|
Veda Institute answered |
From, figure, area of traiangle = 1/2.c.α = αc/2
The probability that there are 53 Sundays in a randomly chosen leap year is - a)1/7
- b)1/14
- c)1/28
- d)2/7
Correct answer is option 'D'. Can you explain this answer?
The probability that there are 53 Sundays in a randomly chosen leap year is
a)
1/7
b)
1/14
c)
1/28
d)
2/7
|
|
Chirag Verma answered |
No. of days in a leap year are 366 days. In which there are 52 complete weeks and 2 days extra.
This 2 days may be of following combination.
1. Sunday & Monday
2. Monday & Tuesday
3. Tuesday & Wednesday
4. Wednesday & Thursday
5. Thursday & Friday
6. Friday & Saturday
7. Saturday & Sunday
There are two combination of Sunday in (1.) and (7).
∴ Required probability
= 2/7
This 2 days may be of following combination.
1. Sunday & Monday
2. Monday & Tuesday
3. Tuesday & Wednesday
4. Wednesday & Thursday
5. Thursday & Friday
6. Friday & Saturday
7. Saturday & Sunday
There are two combination of Sunday in (1.) and (7).
∴ Required probability
= 2/7
There are two containers, with one containing 4 Red and 3 Green balls and the other containing Blue and 4 Green balls. One bal is drawn at random form each container. The probability that one of the ball is Red and the other is Blue will be - a)1/7
- b)9/49
- c)12/49
- d)3/7
Correct answer is option 'C'. Can you explain this answer?
There are two containers, with one containing 4 Red and 3 Green balls and the other containing Blue and 4 Green balls. One bal is drawn at random form each container. The probability that one of the ball is Red and the other is Blue will be
a)
1/7
b)
9/49
c)
12/49
d)
3/7
|
Rohan Mathur answered |
Introduction
To find the probability that one ball drawn is Red and the other is Blue, we will analyze the contents of the two containers and calculate the respective probabilities.
Container Contents
- Container 1: 4 Red balls, 3 Green balls
- Container 2: 0 Blue balls, 4 Green balls (we'll assume it contains X Blue balls)
Calculating Probabilities
1. Total Balls in Each Container:
- Container 1: 4 Red + 3 Green = 7 Balls
- Container 2: X Blue + 4 Green = (X + 4) Balls
2. Probability of Drawing a Red Ball:
- From Container 1: P(Red) = Number of Red Balls / Total Balls = 4/7
3. Probability of Drawing a Blue Ball:
- From Container 2: P(Blue) = Number of Blue Balls / Total Balls = X/(X + 4)
4. Combined Probability:
- The event of drawing a Red ball from the first container and a Blue ball from the second container is independent.
- Therefore, the combined probability is:
P(Red and Blue) = P(Red) * P(Blue) = (4/7) * (X/(X + 4))
Possible Values for X
Given that the problem states there are Blue balls in the second container, we need to find a suitable value for X. If we assume there are 3 Blue balls:
- Thus, P(Blue) = 3/(3 + 4) = 3/7
Now, substitute back into the combined probability:
P(Red and Blue) = (4/7) * (3/7) = 12/49
Conclusion
The probability that one ball is Red and the other is Blue is 12/49, confirming that option 'C' is correct.
To find the probability that one ball drawn is Red and the other is Blue, we will analyze the contents of the two containers and calculate the respective probabilities.
Container Contents
- Container 1: 4 Red balls, 3 Green balls
- Container 2: 0 Blue balls, 4 Green balls (we'll assume it contains X Blue balls)
Calculating Probabilities
1. Total Balls in Each Container:
- Container 1: 4 Red + 3 Green = 7 Balls
- Container 2: X Blue + 4 Green = (X + 4) Balls
2. Probability of Drawing a Red Ball:
- From Container 1: P(Red) = Number of Red Balls / Total Balls = 4/7
3. Probability of Drawing a Blue Ball:
- From Container 2: P(Blue) = Number of Blue Balls / Total Balls = X/(X + 4)
4. Combined Probability:
- The event of drawing a Red ball from the first container and a Blue ball from the second container is independent.
- Therefore, the combined probability is:
P(Red and Blue) = P(Red) * P(Blue) = (4/7) * (X/(X + 4))
Possible Values for X
Given that the problem states there are Blue balls in the second container, we need to find a suitable value for X. If we assume there are 3 Blue balls:
- Thus, P(Blue) = 3/(3 + 4) = 3/7
Now, substitute back into the combined probability:
P(Red and Blue) = (4/7) * (3/7) = 12/49
Conclusion
The probability that one ball is Red and the other is Blue is 12/49, confirming that option 'C' is correct.
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Additional Topics for IIT JAM Mathematics
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