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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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