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Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.
Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(AE1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(AE2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is:
P(E1  A) = P(E1)P(A  E1)/( P(E1)P(A│E1)+P(E2)P(A  E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.
Bayes theorem formula is P(A  B) =
The formula provides relationship between P(A  B) and P(B  A). It is mainly derived from conditional probability formula P(A  B) and P(B  A). Where,
Previous probabilities in Bayes Theorem that are changed with the new available information are called _____
In Bayesian statistics, we calculate new probability after information becomes available due to new events and this is known as Posterior Probability. There is no term like Independent probabilities and Dependent probabilities, there are only independent events and dependent events. Interior probabilities represent probabilities of the intersection between two events.
Bag 1 contains 3 red and 5 black balls while another Bag 2 contains 4 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag 2.
Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a red ball.
P(E1) = P(E2) = 1/2.
Also, P(A  E1) = P(drawing a red ball from Bag 1) = 3/8.
And P(A  E2) = P(drawing a red ball from Bag 2) = 4/10.
The probability of drawing a ball from bag 2, being given that it is red is P(E2  A).
By using Bayes’ theorem,
P(E2  A) = P(E2)P(A  E2)/( P(E1)P(A│E1)+P(E2)P(A  E2))
= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.
_____ is the complement of the angle between the line L and a normal line to the plane π.
The angle between a line and a plane is the complement of the angle between the line L and a normal line to the plane π. If θ is the angle between line whose ratios are a_{1}, b_{1}, c_{1} and the plane ax + by + cz + d = 0 then sin θ = .
Find the angle between the planes x + 2y + 3z + 1 = 0 and (4, 1, 7).
Angle between a plane and a line sin θ =
sin θ = – 0.49
θ = sin1( 0.49)
θ = – 29.34
What is the plane equation involved in the formula sinθ =?
The angle between a line and a plane is the complement of the angle between the line L and a normal line to the plane π. If θ is the angle between line whose ratios are a_{1}, b_{1}, c_{1 }and the plane ax + by + cz + d = 0 then sin θ =
What is the relation between the plane ax + by + cz + d = 0 and a_{1}, b_{1}, c_{1} the direction ratios of a line, if the plane and line are parallel to each other?
The relation between the plane ax + by +cz + d = 0 and a_{1}, b_{1}, c_{1} the direction ratios of a line, if the plane and line are parallel to each other is a_{1}a + b_{1}b + c_{1}c = 0.
The condition a_{1}a + b_{1}b + c_{1}c = 0 is for a plane and a line are _____ to each other.
The relation between the plane ax + by +cz + d = 0 and a_{1}, b_{1}, c_{1} the direction ratios of a line, if the plane and line are parallel to each other is a_{1}a + b_{1}b + c_{1}c = 0.
Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).
Angle between a plane and a line sin θ =
sinθ = 0.49
θ = sin1(0.49)
θ = 29.34
The plane 5x + y + kz + 1 = 0 and directional ratios of a line (3, 1, 1) are parallel, find k.
The condition for a plane and a line are parallel to each other is a_{1}a + b_{1}b + c_{1}c = 0.
5(3) + 1(1) + k(1) = 0
K(1) = 14
K = 14
Find k for the given plane x + 2y + kz + 2 = 0 and directional ratios of a line (8, 3, 2), if they are parallel to each other.
The condition for a plane and a line are parallel to each other is a_{1}a + b_{1}b + c_{1}c = 0.
8(1) + 3(2) + 2(k) = 0
2(k) = 14
k = 7
If θ is the angle between line whose ratios are a1, b1, c1 and the plane ax + by + cz + d = 0 then
cos θ =
A mathematical symbol θ is used to find the angle between line and a normal line to the plane π along with a trigonometric function called sine. Hence, the formula
sin θ =
Which trigonometric function is used to find the angle between a line and a plane?
The trigonometric function is used to find the angle between a line and a plane is sine. If θ is the angle between line whose ratios are a_{1}, b_{1}, c_{1} and the plane ax + by + cz + d = 0 then sin θ = .
A plane and a line having an angle of 90 degrees between them are called _____
A plane and A line which are perpendicular to each other or a plane and a line having an angle 90 degrees between them are called orthogonal. θ is equal to 90 degrees in sin θ = .
The condition a/a1 = b/b1 = c/c1 is for a plane and a line are _____ to each other.
θ = 90 degrees
The relation between the plane ax + by + cz + d = 0 and a_{1}, b_{1}, c_{1} the direction ratios of a line, if the plane and line are perpendicular to each other is a/a_{,} = b/b_{1} = c/c_{1}.
Find the angle between x + 2y + 7z + 2 = 0 and (2, 4, 6).
Angle between a plane and a line sin θ = .
sinθ = 0.92
θ = 66.92
Find the angle between the planes 5x + 2y + 3z + 1 = 0 and (1, 1, 2).
Angle between a plane and a line sin θ = .
sinθ = 0.06
θ = 3.43
_____ is the angle between the normals to two planes.
The angle between the normals to two planes is called the angle between the planes. A trigonometric identity, cosine is used to find the angle called ‘θ’ between two planes.
What is the formula to find the angle between the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0?
The formula to find the angle between the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 is cos θ = .
θ is the angle between the normal of two planes.
Find s for the given planes 2x + 2y + sz + 2 = 0 and 3x + y + z – 2 = 0, if they are perpendicular to each other.
If their normals are perpendicular to each other then a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0.
2(3) + 2(1) + s(1) = 0
s(1) =  8
k =  8
What is the relation between the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c2_{1}z + d_{2} = 0, if their normal are perpendicular to each other?
θ = 90 degrees ⇒ cos θ
a_{1}a_{2} + b_{1}b_{2} – c_{1}c_{2} = 0
Relation between the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c2_{1}z + d_{2} = 0, if their normal are perpendicular to each other is a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0.
The planes 5x + y + 3z + 1 = 0 and x + y – kz + 6 = 0 are orthogonal, find k.
Relation between the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c2_{1}z + d_{2} = 0, if their normal are perpendicular to each other is a1a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0.
5(1) + 1(1) + 3(k) = 0
3k = 6
K = 2
Find k for the given planes x + 2y + kz + 2 = 0 and 3x + 4y – z + 2 = 0, if they are perpendicular to each other.
Relation between the the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c2_{1}z + d_{2} = 0, if their normal are perpendicular to each other is a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0.
1(3) + 2(4) + k(1) = 0
k(1) = 11
k = 11
Which of the following is not the correct formula for representing a plane?
, ax + by + cz = d, lx + my + nz = d are the various ways of representing a plane.
is the vector equation of the plane, where n^ is the unit vector normal to the plane.
ax + by + cz = d, lx + my + nz = d are the Cartesian equation of the plane in the normal form where, a, b, c are the direction ratios and l, m, n are the direction cosines of the normal to the plane respectively.
Find the vector equation of the plane which is at a distance of 7/√38 from the origin and the normal vector from origin is ?
Let
Hence, the required equation of the plane is
Given that the equation of the plane is = 4.
We know that,
∴
⇒ 2x + y  z = 4 is the Cartesian equation of the plane.
Find the distance of the plane 3x + 4y  5z  7=0.
From the given equation, the direction ratios of the normal to the plane are 3, 4, 5; the direction cosines are
, i.e. 3√50,4√50,−5√50
Dividing the equation throughout by √50, we get
The above equation is in the form of lx + my + nz = d, where d is the distance of the plane from the origin. So, the distance of the plane from the origin is 7√50.
Find the equation of the plane passing through the three points (2,2,0), (1,2,1), (1,2,2).
Let,
The vector equation of the plane passing through three points is given by
= 0
= 0
Find the Cartesian equation of the plane passing through the point (3,2,3) and the normal to the plane is ?
The position vector of the point (3,2,3) is and the normal vector perpendicular to the plane is
Therefore, the vector equation of the plane is given by = 0
Hence,
= 0
4(x  3)2(y2) + 5(z + 3) = 0
4x  2y + 5z + 7=0.
Given that the equation of the plane is
We know that,
∴ = 12
⇒ (λ + 2μ)x + (2λ  μ)y + (3λ  2μ)z = 12 is the Cartesian equation of the plane.
Find the Cartesian equation of the plane passing through the point (3,2,3) and the normal to the plane is
The position vector of the point (3,2,3) is and the normal vector perpendicular to the plane is
Therefore, the vector equation of the plane is given by = 0
Hence, = 0
4(x  3)  2(y  2) + 5(z + 3) = 0
4x  2y + 5z + 7 = 0
To evaluate:
=
=
Explanation: Given that, and
Calculating the vector product, we get
=
=
Given that, and
Calculating the vector product, we get
=
=
Find the magnitude of a⃗ and b⃗ which are having the same magnitude and such that the angle between them is 60° and their scalar product is 1/4.
Given that: a)
b) θ = 60° and
c) = 14
∴ = 1/4
=  = 1/4
⇒ ^{ }= 1/4 x 1/2
∴
Consider
=
=
We know that, and
∴
=
Hence,
Direction of λ and is same if value λ is positive as it gives it a direction which is positive in nature. If the value of λ is negative then the direction of the result after multiplication becomes in opposite direction. Whereas the value of the product vector becomes zero if value of λ is 0.
λ times the magnitude of vector is denoted as ______
λ times the magnitude of vector is denoted as
As we know that the magnitude of vector is denoted by , if we multiply the magnitude of vector with magnitude of λ we get .
As both the vectors are equal hence, we can equate their constants and get the value of x, y and z. Now we equate the coefficients of of both the equations and get the values
x = 2, y = 2, z = 1.
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