WAEC SSCE - Further Mathematics - 2021

A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.

A. 12/91 B. C. D.

Bayanin Amsa

Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)

Bayanin Amsa

To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).

Bayanin Amsa

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the distance travelled in the third second.

Bayanin Amsa

Given that F\(^1\)(x) = x\(^3\) √x, find f(x)

Bayanin Amsa

In how many ways can 8 persons be seated on a bench if only three seats are available?

Bayanin Amsa

The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)

Bayanin Amsa

Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)

Bayanin Amsa

To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R

Bayanin Amsa

If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)

Bayanin Amsa

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the maximum height reached.

Bayanin Amsa

The table shows the distribution of marks obtained by some students in a test

What is the upper class boundary of the upper quartile class?

Bayanin Amsa

The first term of an AP is 4 and the sum of the first three terms is 18. Find the product of the first three terms

Bayanin Amsa

Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)

Bayanin Amsa

A fair die is tossed 60 times and the results are recorded in the table

Find the probability of obtaining a prime number.

Bayanin Amsa

A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

Consider the following statements:

X: Benita is polite

y: Benita is neat

z: Benita is intelligent

Which of the following symbolizes the statement: "Benita is neat if and only if she is neither polite nor intelligent"?

Bayanin Amsa

A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?

Bayanin Amsa

The table shows the distribution of marks obtained by some students in a test

Find the modal class mark.

Bayanin Amsa

To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N

Bayanin Amsa

If 2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

Bayanin Amsa

For what range of values of x is x\(^2\) - 2x - 3 ≤ 0

Bayanin Amsa

To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

Given that P = { x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3} and Q = { x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P n Q.

Bayanin Amsa

If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find  cos (x + y)

Bayanin Amsa

A body of mass 15kg is placed on a smooth plane which is inclined at 60° to the horizontal. If the box is at rest,

calculate the normal reaction to the plane. [ Take g = 10m/s\(^2\) ]

Bayanin Amsa

To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

If 2i +pj and 4i -2j are perpendicular, find the value of p.

Bayanin Amsa

To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p

Bayanin Amsa

Find the inverse of  \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)

Bayanin Amsa

g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4

Bayanin Amsa

A binary operation * is defined on the set of real numbers, R, by

P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2

Bayanin Amsa

The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\) 

Bayanin Amsa

Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0. 

Bayanin Amsa

To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P. 

Bayanin Amsa

Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2. 

Bayanin Amsa

Given that f: x --> x\(^2\) - x + 1 is defined on the Set Q = { x : 0 ≤ x < 20, x is a multiple of 5}. find the set of range of F.

Bayanin Amsa

For what value of k is 4x\(^2\) - 12x + k, a perfect square?

Bayanin Amsa

If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.

Bayanin Amsa

Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

Bayanin Amsa

Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7).

Bayanin Amsa

P and Q are two linear transformations in the X-Y plane defined by

P: (x, y) → (-3x + 6y, 4x + y) and

Q: (x, y) → (2x-3y, -4x - 6y).

(a) Write down the matrices of P and Q. (b) What is the image of (-2,-3) under the transformation Q?

(c) Obtain a single transformation representing the transformation Q followed by P.

(d) Find the image of (1,4) when transformed by Q followed by P.

(e) Find the image P\(^1\) of the point (-√2,2√2) under an anticlockwise rotation of 225° about the origin.

Bayanin Amsa

(a) The matrix representation of a linear transformation can be obtained by writing the image of the standard basis vectors (1,0) and (0,1). The matrix of P is given by: P = [[-3, 6], [4, 1]] Similarly, the matrix of Q is given by: Q = [[2, -3], [-4, -6]] (b) To find the image of a point (-2,-3) under the transformation Q, we first write the point as a column vector: (-2,-3) = [-2, -3]^T Next, we multiply the matrix Q with the vector: Q * [-2, -3]^T = [2, -3] * [-2, -3]^T = [-12, 18]^T So the image of (-2,-3) under the transformation Q is (-12, 18). (c) To find the single transformation representing the transformation Q followed by P, we multiply the matrices Q and P: Q * P = [[2, -3], [-4, -6]] * [[-3, 6], [4, 1]] = [[36, -18], [-24, -30]] So the single transformation representing the transformation Q followed by P is given by the matrix [[36, -18], [-24, -30]]. (d) To find the image of (1,4) when transformed by Q followed by P, we first write the point as a column vector: (1, 4) = [1, 4]^T Next, we multiply the matrix representing Q followed by P with the vector: [[36, -18], [-24, -30]] * [1, 4]^T = [36, -18] * 1 + [-24, -30] * 4 = [36, -18] + [-96, -120] = [-60, -138]^T So the image of (1,4) when transformed by Q followed by P is (-60, -138). (e) To find the image P^1 of the point (-√2,√2) under an anticlockwise rotation of 225° about the origin, we can first rotate the point by 225° in the counterclockwise direction and then apply the transformation P. The counterclockwise rotation of 225° can be represented by the matrix: R = [[cos(225), -sin(225)], [sin(225), cos(225)]] = [[-√2/2, √2/2], [-√2/2, -√2/2]] Next, we multiply the matrix R with the vector representing the point (-√2,√2): R * [-√2, √2]^T = [[-√2/2, √2/2], [-√2/2, -√2/2]] * [-√2, √2]^T = [-√2/2 * -√2 + √2/2 * √2, -√2/2 * -√2 - √2/2 * √2]^T = [√2, -√2]^T So the image of the point (-√2,√2) under the counterclockwise rotation of 225° is (√2, -√2). Finally, to find

\(^{5y}{C}_2\) = 190, find the value of y

Bayanin Amsa

We can use the formula for combinations to solve for y:

^5yC₂ = ^(5y)! / _(2!(5y-2)!) = 190

Expanding the factorials:

^(5y)(5y-1) / 2 = 190

Solving for y:

(5y)(5y-1) = 380

25y^2 - 5y = 380

25y^2 - 5y - 380 = 0

Using the quadratic formula:

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))

y = (5 ± √(25 + 90000))/(50)

y = (5 ± √(90025))/(50)

Taking the positive root:

y = (5 + 300)/(50)

y = 305/50

y = 6.1

So the value of y is approximately 6.1.

Given that x =  \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors

Bayanin Amsa

x =  (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$

Changing x and y to the form xi+yj

we have:

x= -4i + 3j and y = -9i - 15j

using:

cosØ = xy|x|ly| $\frac{𝑥𝑦}{|𝑥|𝑙𝑦|}$

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(42 ${}^2$ +32 ${}^2$) =√(16 +9) = √25 = 5

|y| = √(-9)2 ${}^2$ + (-15)2 ${}^2$ = √(81 +225) = √306

cosØ =−815√306 $\frac{-81}{5\surd 306}$

= −9√306170 $\frac{-9\surd 306}{170}$

17.49∗9170 $\frac{17.49\ast 9}{170}$

Ø = cos−1 ${}^{-1}$(-0.1029);

Ø = 95.91°

Ø = 96°

(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.

(i) How many days will it take the jogger to reach a distance of 15km in training?

(ii) Calculate the total distance he would have run in the training.

(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.

Bayanin Amsa

(a)

(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.

Let's call the number of days it takes to reach this distance "n."

On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.

In general, on the nth day, he runs 500 + 250(n-1) meters.

So we can set up an equation:

500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000

Simplifying, we get:

250n^2 + 250n - 15000 = 0

Dividing both sides by 250:

n^2 + n - 60 = 0

This equation can be factored as:

(n + 6)(n - 10) = 0

Since we're looking for a positive value for n, the answer is n = 10.

So it will take the jogger 10 days to reach a distance of 15km in training.

(ii) We can use the formula for the sum of a geometric series:

S = a(1 - r^n)/(1 - r)

where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.

In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.

We also know that the sum to infinity is 25/2, so S = 25/2.

Plugging in these values and solving for r:

25/2 = (-1500)(1 - r^10)/(1 - r)

r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)

So the common ratio is 1/2.

The total distance the jogger would have run in the training is the sum of the terms of the geometric series:

S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.

(b)

We know that the second term of the GP is -3, so we can write the first two terms as:

a, ar

where ar = -3.

We also know that the sum to infinity is 25/2, so we can use the formula:

S = a/(1 - r)

25/2 = a/(1 - r)

a = 25/2 - 25r/2

Substituting this value of a into the equation ar = -3:

(25/2 - 25r/2)r = -3

Simplifying:

25r^2 - 50r + 6 = 0

We can solve for r using the quadratic formula:

r = (50

Given that (p + 1/2√3)(1 - √3)\(^2\) = 3- √3,

 find x the value of p.

Bayanin Amsa

To solve for p in the given equation, we can start by simplifying the left-hand side of the equation using the identity (a + b)(a - b) = a² - b²:

(p + 1/2√3)(1 - √3)² = (p + 1/2√3)(1 - 2√3 + 3)
          = (p + 1/2√3)(4 - 2√3)
          = 4p - 2p√3 + √3/2 - 1/2
          = (4p - 1/2) - (2p√3 - √3/2)

Now we can set this expression equal to the right-hand side of the equation and solve for p:

4p - 1/2 - (2p√3 - √3/2) = 3 - √3
4p - 2p√3 = 7/2 - √3/2
2p(2 - √3) = 7/2 - √3/2
p = (7/4 - √3/4)/(2 - √3)

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is 2 + √3:

p = [(7/4 - √3/4)(2 + √3)]/[(2 - √3)(2 + √3)]
p = (7/2 + √3/2 - 2√3/4 - √3/4)/(4 - 3)
p = (7/2 - 3√3/4)/1
p = 14/4 - 3√3/4
p = (7 - √3)/2

Therefore, the value of p that satisfies the given equation is (7 - √3)/2.

The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find:

(a) the coordinates of M;

(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.

Bayanin Amsa

To solve this problem, we first need to understand what a parallelogram is. A parallelogram is a four-sided shape with opposite sides that are parallel to each other. This means that if we draw lines from opposite corners of the parallelogram, these lines will intersect at the midpoint of the two sides they are drawn from.

Now let's look at the given position vectors of P, Q, and R. We can use these to find the position vector of M, which is opposite to P in the parallelogram. To do this, we need to use the fact that the position vector of the midpoint of a line segment is the average of the position vectors of the two endpoints. So, the position vector of M is:

¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ =  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ - (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

MR¯¯¯¯¯¯¯¯¯ $\overline{𝑀𝑅}$ =  (−52) $\left(\begin{array}{c}-5\\ 2\end{array}\right)$ - (xy) $\left(\begin{array}{c}𝑥\\ 𝑦\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

 (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)

(b) PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$ = OM¯¯¯¯¯¯¯¯¯ $\overline{𝑂𝑀}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=   (−2−6) $\left(\begin{array}{c}-2\\ -6\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−6−1) $\left(\begin{array}{c}-6\\ -1\end{array}\right)$

    PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ = OQ¯¯¯¯¯¯¯¯ $\overline{𝑂𝑄}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

|PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$| = √(-62 ${}^2$ + [-12 ${}^2$]) = √37

|PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$| = √(-32 ${}^2$ + 82 ${}^2$) = √73

cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯| $\frac{𝑃𝑀.𝑃𝑄}{|\overline{𝑃𝑀}||\overline{𝑃𝑄}|}$

cosØ = [−6i−j].[−3i+8j]√37.√73 $\frac{[-6𝑖-𝑗].[-3𝑖+8𝑗]}{\surd 37.\surd 73}$ = 10√2710 $\frac{10}{\surd 2710}$

⇒ Ø = cos−1 ${}^{-1}$ 10√2710 $\frac{10}{\surd 2710}$

: Ø = 78.91° ≈ 79°

A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.

Calculate the probability of picking:

(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.

Bayanin Amsa

Total bulbs:

n(Red)=5, n(B) = 7, n(G) = 4

(5+7+4) = 16

p(R)= 5/16, p(B) = 7/16, p(G) = 4/16

(a) p(All same colour of bulbs) 

= p(RR) Or p(BB) or p(GG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 716 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 615 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 416 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 315

= 112 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 740 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

=  10+21+6120 $\frac{<br>\mathrm{<br>}}{}$ =  37120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

 (b) All different colors = 1-p (AIl the same colour) =

1 - 37120 $\frac{37}{120}$ = 83120 $\frac{83}{120}$

(c) p(At least one red)  = p(RR) + p(RB) + p(RG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 715 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

= 112 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 748 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  112 $\frac{\mathrm{<br>}}{}$

8+748 $\frac{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}{\mathrm{<br>}}$ = 1548 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$