For independent events A and B, the probability of both occurring is given by: P(A ∩ B) = P(A) × P(B) This is the fundamental property of independent events in probability. So the correct answer is P(A). P(B) Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-Read more
For independent events A and B, the probability of both occurring is given by:
P(A ∩ B) = P(A) × P(B)
This is the fundamental property of independent events in probability.
Given: P(E) = 0.3 P(E ∪ F) = 0.5 Since E and F are independent events, we use the formula: P(E ∪ F) = P(E) + P(F) - P(E ∩ F) Given that E and F are independent: P(E ∩ F) = P(E) × P(F) Substitute the values: 0.5 = 0.3 + P(F) - (0.3 × P(F)) 0.5 = 0.3 + P(F) - 0.3P(F) 0.2 = P(F) - 0.3P(F) 0.2 = 0.7P(F)Read more
Given:
P(E) = 0.3
P(E ∪ F) = 0.5
Since E and F are independent events, we use the formula:
P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
Given that E and F are independent:
P(E ∩ F) = P(E) × P(F)
For independent events A and B the conditional probability follows: P(A | B) = P(A ∩ B) / P(B). Since the events A and B are independent, we recall that: P(A ∩ B) = P(A) × P(B). Thus P(A | B) = P(A) (P(B))/(P(B)), canceling (P(B), we have ) P(A) So the correct answer is P(A) Click here for more: httRead more
For independent events A and B the conditional probability follows: P(A | B) = P(A ∩ B) / P(B). Since the events A and B are independent, we recall that: P(A ∩ B) = P(A) × P(B). Thus P(A | B) = P(A) (P(B))/(P(B)), canceling (P(B), we have ) P(A)
A second-order determinant is of the form: | a b | | c d | Each element (a, b, c, d) can be either 0 or 1. Total possible cases = 2⁴ = 16 The determinant value is given by: Det = (a × d) - (b × c) For the determinant to be **non-positive**, we need: (a × d) - (b × c) ≤ 0 Now, let's count the favorabRead more
A second-order determinant is of the form:
| a b |
| c d |
Each element (a, b, c, d) can be either 0 or 1.
Total possible cases = 2⁴ = 16
The determinant value is given by:
Det = (a × d) – (b × c)
For the determinant to be **non-positive**, we need:
(a × d) – (b × c) ≤ 0
A fair coin is tossed 6 times. Let X be the product of the number of heads (H) and the number of tails (T). Since each toss results in either a head or a tail: Total tosses = 6 → H + T = 6 Possible values of (H, T) pairs: (0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0) Now, determine X = H × T: - (0Read more
A fair coin is tossed 6 times.
Let X be the product of the number of heads (H) and the number of tails (T).
Since each toss results in either a head or a tail:
Total tosses = 6 → H + T = 6
Possible values of (H, T) pairs:
(0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)
Given: P(A) = 0.2 P(B) = 0.4 P(A ∪ B) = 0.5 Using the formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) Substituting the values: 0.5 = 0.2 + 0.4 - P(A ∩ B) 0.5 = 0.6 - P(A ∩ B) P(A ∩ B) = 0.1 Now, using the conditional probability formula: P(A | B) = P(A ∩ B) / P(B) P(A | B) = 0.1 / 0.4 P(A | B) = 0.25 SoRead more
Given:
P(A) = 0.2
P(B) = 0.4
P(A ∪ B) = 0.5
Using the formula:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Substituting the values:
0.5 = 0.2 + 0.4 – P(A ∩ B)
0.5 = 0.6 – P(A ∩ B)
P(A ∩ B) = 0.1
Now, using the conditional probability formula:
P(A | B) = P(A ∩ B) / P(B)
P(A | B) = 0.1 / 0.4
P(A | B) = 0.25
As follows from the definition, if events A and B are independent, the correct statement is: - A and B are independent For independent events, the conditional probability P(A | B) = P(A), and P(B | A) = P(B), because the occurrence of one event does not affect the probability of the other. The stateRead more
As follows from the definition, if events A and B are independent, the correct statement is:
– A and B are independent
For independent events, the conditional probability P(A | B) = P(A), and P(B | A) = P(B), because the occurrence of one event does not affect the probability of the other.
The statement “P(A | B) + P(A | B) = 1” is wrong and does not generally hold.
Therefore, the right answer is A and B are independent.
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass ofRead more
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass of the system.
Here the rod is of length 1.4 m. The point masses are 0.3 kg and 0.7 kg located at the opposite ends of this rod. It is found that the center of mass is placed at a distance of about 0.98 m from the 0.3 kg mass. So to minimize the work done in the rotation, this axis of rotation should pass through this point.
Positioning the axis at the center of mass minimizes the rotational inertia of the system, making the rotation more efficient. This is one of the approaches in physics to make calculations easier and reduce the energy required to start rotational motion.
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocityRead more
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocity of 14 m/s towards the minor block, given that the little block is assumed to be moving on a horizontal frictionless ground.
It is the average position of the entire mass in the system weighted by their respective velocities. The velocity of the center of mass for this system would depend on the contributions from each block’s mass and its respective velocity. In this case, because the light block has no movement at all it contributes nothing to the velocity of the center of mass. Its movement is dominated by that of the heavier block since the former has more mass and higher speed.
By adding both blocks’ masses and velocities together, the center of mass’s velocity is found to be 10 m/s. This represents the motion of the system in a frictionless surface. The center of mass’s velocity is only changed when acted on by an outside force according to the principles of physics.
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with aRead more
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with a mass of m/4 and the other with a mass of 3m/4.
Given after the explosion that the smaller fragment is located at y = +15 cm, use the idea that the overall center of mass for the system must remain at the same vertical position as before the explosion, which was at y = 0.
The position of the center of mass is determined by the masses and their respective positions. Since the smaller fragment is at y = +15 cm, the position of the larger fragment must be such that it balances this position according to their mass ratio. By conservation, we determine that the larger fragment must be positioned at y = -5 cm. This negative position denotes that it lies below the initial center of mass level, making sure the global system maintains the center of mass position after detonation.
If the events A and B are independent, then P(A ∩ B) is equal to
For independent events A and B, the probability of both occurring is given by: P(A ∩ B) = P(A) × P(B) This is the fundamental property of independent events in probability. So the correct answer is P(A). P(B) Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-Read more
For independent events A and B, the probability of both occurring is given by:
P(A ∩ B) = P(A) × P(B)
This is the fundamental property of independent events in probability.
So the correct answer is P(A). P(B)
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Two events E and F are independent. If P(E) = 0.3, P(E U F) = 0.5, then P(E/F) – P(F/E) equals
Given: P(E) = 0.3 P(E ∪ F) = 0.5 Since E and F are independent events, we use the formula: P(E ∪ F) = P(E) + P(F) - P(E ∩ F) Given that E and F are independent: P(E ∩ F) = P(E) × P(F) Substitute the values: 0.5 = 0.3 + P(F) - (0.3 × P(F)) 0.5 = 0.3 + P(F) - 0.3P(F) 0.2 = P(F) - 0.3P(F) 0.2 = 0.7P(F)Read more
Given:
P(E) = 0.3
P(E ∪ F) = 0.5
Since E and F are independent events, we use the formula:
P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
Given that E and F are independent:
P(E ∩ F) = P(E) × P(F)
Substitute the values:
0.5 = 0.3 + P(F) – (0.3 × P(F))
0.5 = 0.3 + P(F) – 0.3P(F)
0.2 = P(F) – 0.3P(F)
0.2 = 0.7P(F)
P(F) = 2/7
Now, conditional probability calculations:
P(E | F) = P(E ∩ F) / P(F)
= (0.3 × 2/7) ÷ (2/7)
= 0.3
P(F | E) = P(E ∩ F) / P(E)
= (0.3 × 2/7) ÷ 0.3
= 2/7
Now, computing the needed difference:
P(E | F) – P(F | E) = 0.3 – 2/7
= 3/10 – 2/7
By taking LCM (70):
= (21/70) – (20/70)
= 1/70
Hence, the correct answer is 1/70
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If A and B are independent events, then P(A/B ) equals
For independent events A and B the conditional probability follows: P(A | B) = P(A ∩ B) / P(B). Since the events A and B are independent, we recall that: P(A ∩ B) = P(A) × P(B). Thus P(A | B) = P(A) (P(B))/(P(B)), canceling (P(B), we have ) P(A) So the correct answer is P(A) Click here for more: httRead more
For independent events A and B the conditional probability follows: P(A | B) = P(A ∩ B) / P(B). Since the events A and B are independent, we recall that: P(A ∩ B) = P(A) × P(B). Thus P(A | B) = P(A) (P(B))/(P(B)), canceling (P(B), we have ) P(A)
So the correct answer is P(A)
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If each element of a second order determinant is either zero or one, the probability that the value of determinant is non-positive is
A second-order determinant is of the form: | a b | | c d | Each element (a, b, c, d) can be either 0 or 1. Total possible cases = 2⁴ = 16 The determinant value is given by: Det = (a × d) - (b × c) For the determinant to be **non-positive**, we need: (a × d) - (b × c) ≤ 0 Now, let's count the favorabRead more
A second-order determinant is of the form:
| a b |
| c d |
Each element (a, b, c, d) can be either 0 or 1.
Total possible cases = 2⁴ = 16
The determinant value is given by:
Det = (a × d) – (b × c)
For the determinant to be **non-positive**, we need:
(a × d) – (b × c) ≤ 0
Now, let’s count the favorable cases:
1. Cases where determinant = 0:
– (a, b, c, d) = (0,0,0,0) → Det = (0×0) – (0×0) = 0
– (0,0,0,1) → (0×1) – (0×0) = 0
– (0,0,1,0) → (0×0) – (0×1) = 0
– (0,1,0,0) → (0×0) – (1×0) = 0
– (1,0,0,0) → (1×0) – (0×0) = 0
– (1,1,1,1) → (1×1) – (1×1) = 0
So, 6 cases where Det = 0.
2. Cases where determinant < 0:
– (0,1,1,0) → (0×0) – (1×1) = -1
– (1,1,0,0) → (1×0) – (1×0) = 0
– (0,0,1,1) → (0×1) – (0×1) = 0
– (1,0,0,1) → (1×1) – (0×0) = 1 (not negative)
So, 5 cases where Det < 0.
Total favorable cases = 6 (Det = 0) + 5 (Det < 0) = 11
Probability = 11/16
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Let X represent the product of the number of heads and the number of tails obtained when a coin is tossed 6 times. Then, the possible values of X are
A fair coin is tossed 6 times. Let X be the product of the number of heads (H) and the number of tails (T). Since each toss results in either a head or a tail: Total tosses = 6 → H + T = 6 Possible values of (H, T) pairs: (0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0) Now, determine X = H × T: - (0Read more
A fair coin is tossed 6 times.
Let X be the product of the number of heads (H) and the number of tails (T).
Since each toss results in either a head or a tail:
Total tosses = 6 → H + T = 6
Possible values of (H, T) pairs:
(0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)
Now, determine X = H × T:
– (0,6) → 0 × 6 = 0 .
– (1,5) → 1 × 5 = 5 .
– (2,4) → 2 × 4 = 8 .
– (3,3) → 3 × 3 = 9 .
– (4,2) → 4 × 2 = 8 .
– (5,1) → 5 × 1 = 5
– (6,0) → 6 × 0 = 0
Possible values of X = {0, 5, 8, 9}
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If A and B are events such that P(A) = 0.2, P(B) = 0.4 and P(A U B) = 0.5, then value of P(A/B) is
Given: P(A) = 0.2 P(B) = 0.4 P(A ∪ B) = 0.5 Using the formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) Substituting the values: 0.5 = 0.2 + 0.4 - P(A ∩ B) 0.5 = 0.6 - P(A ∩ B) P(A ∩ B) = 0.1 Now, using the conditional probability formula: P(A | B) = P(A ∩ B) / P(B) P(A | B) = 0.1 / 0.4 P(A | B) = 0.25 SoRead more
Given:
P(A) = 0.2
P(B) = 0.4
P(A ∪ B) = 0.5
Using the formula:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Substituting the values:
0.5 = 0.2 + 0.4 – P(A ∩ B)
0.5 = 0.6 – P(A ∩ B)
P(A ∩ B) = 0.1
Now, using the conditional probability formula:
P(A | B) = P(A ∩ B) / P(B)
P(A | B) = 0.1 / 0.4
P(A | B) = 0.25
So the correct answer is 0.25
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If A and B are two independent events, then
As follows from the definition, if events A and B are independent, the correct statement is: - A and B are independent For independent events, the conditional probability P(A | B) = P(A), and P(B | A) = P(B), because the occurrence of one event does not affect the probability of the other. The stateRead more
As follows from the definition, if events A and B are independent, the correct statement is:
– A and B are independent
For independent events, the conditional probability P(A | B) = P(A), and P(B | A) = P(B), because the occurrence of one event does not affect the probability of the other.
The statement “P(A | B) + P(A | B) = 1” is wrong and does not generally hold.
Therefore, the right answer is A and B are independent.
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Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass ofRead more
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass of the system.
Here the rod is of length 1.4 m. The point masses are 0.3 kg and 0.7 kg located at the opposite ends of this rod. It is found that the center of mass is placed at a distance of about 0.98 m from the 0.3 kg mass. So to minimize the work done in the rotation, this axis of rotation should pass through this point.
Positioning the axis at the center of mass minimizes the rotational inertia of the system, making the rotation more efficient. This is one of the approaches in physics to make calculations easier and reduce the energy required to start rotational motion.
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See lessTwo blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocityRead more
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocity of 14 m/s towards the minor block, given that the little block is assumed to be moving on a horizontal frictionless ground.
It is the average position of the entire mass in the system weighted by their respective velocities. The velocity of the center of mass for this system would depend on the contributions from each block’s mass and its respective velocity. In this case, because the light block has no movement at all it contributes nothing to the velocity of the center of mass. Its movement is dominated by that of the heavier block since the former has more mass and higher speed.
By adding both blocks’ masses and velocities together, the center of mass’s velocity is found to be 10 m/s. This represents the motion of the system in a frictionless surface. The center of mass’s velocity is only changed when acted on by an outside force according to the principles of physics.
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See lessAn isolated particle of mass m is moving in horizontal in horizontal plane (x-y) along the x-axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with aRead more
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with a mass of m/4 and the other with a mass of 3m/4.
Given after the explosion that the smaller fragment is located at y = +15 cm, use the idea that the overall center of mass for the system must remain at the same vertical position as before the explosion, which was at y = 0.
The position of the center of mass is determined by the masses and their respective positions. Since the smaller fragment is at y = +15 cm, the position of the larger fragment must be such that it balances this position according to their mass ratio. By conservation, we determine that the larger fragment must be positioned at y = -5 cm. This negative position denotes that it lies below the initial center of mass level, making sure the global system maintains the center of mass position after detonation.
Click here for more: – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
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