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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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