Probability
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New answer posted
a month agoContributor-Level 10
sin?¹(x² + 1/3) + cos?¹(x² - 2/3) = x²
The domains of sin?¹ and cos?¹ require:
-1 ≤ x² + 1/3 ≤ 1 ⇒ -4/3 ≤ x² ≤ 2/3. Since x² ≥ 0, we have 0 ≤ x² ≤ 2/3.
-1 ≤ x² - 2/3 ≤ 1 ⇒ -1/3 ≤ x² ≤ 5/3.
The intersection of these domains is 0 ≤ x² ≤ 2/3.
The range of sin?¹ is [-π/2, π/2] and cos?¹ is [0, π].
Let A = sin?¹(x² + 1/3) and B = cos?¹(x² - 2/3).
The equation is A + B = x².
The LHS, A+B, is a sum of angles, while the RHS, x², is in the range [0, 2/3]. This suggests no solution. The provided solution states that LHS = {π}, which is incorrect. A proper analysis would involve checking if any x in
New answer posted
a month agoContributor-Level 10
Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0
For x < 0: f(x) = x(2 - sin(1/x))
For x > 0: f(x) = x(2 - sin(1/x))
The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)
The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)
To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.
New answer posted
a month agoContributor-Level 10
A line passes through (1,3). Its equation is y - 3 = m (x - 1) or y = mx + (3-m).
The angle θ between this line and the line y = 3√2x - 1 (with slope m? = 3√2) is given by tanθ = √2.
tanθ = | (m - m? )/ (1 + m*m? )|
√2 = | (m - 3√2) / (1 + m*3√2)|
This gives two cases:
Case 1 (+ve):
√2 = (m - 3√2) / (1 + 3√2m)
√2 (1 + 3√2m) = m - 3√2
√2 + 6m = m - 3√2
5m = -4√2
m = -4√2 / 5
Case 2 (-ve):
-√2 = (m - 3√2) / (1 + 3√2m)
-√2 (1 + 3√2m) = m - 3√2
-√2 - 6m = m - 3√2
7m = 2√2
m = 2√2 / 7
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
P (at least 2 show 3 or 5) =? C? (2/6)² (4/6)² +? C? (2/6)³ (4/6)¹ +? C? (2/6)?
= (384+128+16)/6? = 11/27
n=27
∴ expectation of number of times = np
= 27 ⋅ (11/27) = 11
New answer posted
a month agoContributor-Level 10
We have, 1 - (probability of all shots result in failure) > 1/4
? 1 - (9/10)? > 1/4
? 3/4 > (9/10)? ? n? 3 > (9/10)? ? n? 3
New answer posted
a month agoContributor-Level 9
P (A∪B∪C) = P (A) + P (B) + P (C) – P (A∩B) – P (B∩C) – P (C∩A) + P (A∩B∩C)
Given relations lead to: α = 1.4 – P (A∩B) – β ⇒ α + β = 1.4 - P (A∩B)
Again, from P (A∪B) = P (A) + P (B) – P (A∩B), and given values, it is found that P (A∩B) = 0.2.
From (1) and (2), α = 1.2 – β.
Now given 0.85 ≤ α ≤ 0.95
⇒ 0.85 ≤ 1.2 – β ≤ 0.95
⇒ -0.35 ≤ -β ≤ -0.25
⇒ 0.25 ≤ β ≤ 0.35, so β ∈ [0.25, 0.35]
New answer posted
a month agoContributor-Level 10
sum 6 → (1,5), (5,1), (3,3), (2,4), (4,2)
sum 7 → (1,6), (6,1), (5,2), (2,5), (3,4), (4,3)
= P (A) + P (? ) · P (B) · P (A) + P (? )P (B)P (? ) · P (B) · P (A) + …
This is infinite G.P. with common ratio P (? ) * P (B)
Probability of A wins = P (A) / (1 - P (? )P (B? )
= (5/36) / (1 - (31/36)* (30/36) = 30/61
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