Class 10 Board Exam Math

Cheatsheet Content

### 1. Real Numbers - **Euclid's Division Lemma:** Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \le r ### 2. Polynomials - **General Form:** $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$, where $a_n \ne 0$. - **Degree of a Polynomial:** The highest power of the variable in a polynomial. - Linear: Degree 1 ($ax+b$) - Quadratic: Degree 2 ($ax^2+bx+c$) - Cubic: Degree 3 ($ax^3+bx^2+cx+d$) - **Zeros of a Polynomial:** The values of 'x' for which $P(x)=0$. Graphically, these are the x-intercepts. - **Relationship between Zeros and Coefficients:** - **For a quadratic polynomial $ax^2+bx+c$ with zeros $\alpha, \beta$:** - Sum of zeros: $\alpha + \beta = -b/a$ - Product of zeros: $\alpha \beta = c/a$ - **For a cubic polynomial $ax^3+bx^2+cx+d$ with zeros $\alpha, \beta, \gamma$:** - Sum of zeros: $\alpha + \beta + \gamma = -b/a$ - Sum of products of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$ - Product of zeros: $\alpha\beta\gamma = -d/a$ - **Division Algorithm for Polynomials:** If $P(x)$ and $G(x)$ are two polynomials with $G(x) \ne 0$, then we can find polynomials $Q(x)$ and $R(x)$ such that $P(x) = G(x) \cdot Q(x) + R(x)$, where $R(x) = 0$ or $\text{degree of } R(x) ### 3. Pair of Linear Equations in Two Variables - **General Form:** $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. - **Graphical Representation and Conditions for Solvability:** | Condition | Graphical Representation | Algebraic Interpretation | |-----------------------------------------|--------------------------|------------------------------| | $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$ | Intersecting Lines | Unique solution (Consistent) | | $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ | Coincident Lines | Infinitely many solutions (Consistent and Dependent) | | $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$ | Parallel Lines | No solution (Inconsistent) | - **Methods of Solving:** - **Substitution Method:** Express one variable in terms of the other from one equation and substitute into the second equation. - **Elimination Method:** Multiply equations by suitable non-zero constants to make the coefficients of one variable equal, then add or subtract the equations. - **Cross-Multiplication Method:** For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$: $$ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} $$ **Example Question:** Solve the following pair of linear equations by the elimination method: $x + y = 5$ (1) $2x - 3y = 4$ (2) **Solution:** Multiply equation (1) by 3: $3(x + y) = 3(5) \implies 3x + 3y = 15$ (3) Add equation (2) and (3): $(2x - 3y) + (3x + 3y) = 4 + 15$ $5x = 19 \implies x = \frac{19}{5}$ Substitute $x = \frac{19}{5}$ into equation (1): $\frac{19}{5} + y = 5 \implies y = 5 - \frac{19}{5} = \frac{25 - 19}{5} = \frac{6}{5}$ Thus, the solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$. ### 4. Quadratic Equations - **Standard Form:** $ax^2 + bx + c = 0$, where $a \ne 0$. - **Methods of Solving:** - **Factorization (Splitting the Middle Term):** Find two numbers whose sum is $b$ and product is $ac$, then rewrite $bx$ using these numbers. - **Completing the Square:** Transform the equation into the form $(x+k)^2 = d$. - **Quadratic Formula:** $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ - **Discriminant ($D$):** $D = b^2 - 4ac$. Determines the nature of roots. - $D > 0$: Two distinct real roots. - $D = 0$: Two equal real roots (coincident roots). - $D 0$, there are two distinct real roots. Using the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-5) \pm \sqrt{1}}{2(2)} = \frac{5 \pm 1}{4}$. So, $x_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$ and $x_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$. The roots are $3/2$ and $1$. ### 5. Arithmetic Progressions (AP) - **General Form:** A sequence of numbers where the difference between consecutive terms is constant. $a, a+d, a+2d, ...$ - $a$: first term - $d$: common difference - **$n^{th}$ Term (General Term):** $a_n = a + (n-1)d$ - **Sum of First $n$ Terms ($S_n$):** - $S_n = \frac{n}{2}[2a + (n-1)d]$ - $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term ($a_n$). - **Property:** If $a, b, c$ are in AP, then $2b = a+c$. **Example Question:** Find the $10^{th}$ term of the AP: $2, 7, 12, ...$ **Solution:** Here, the first term $a=2$. The common difference $d = 7 - 2 = 5$. To find the $10^{th}$ term, we use the formula $a_n = a + (n-1)d$. For $n=10$, $a_{10} = 2 + (10-1)5 = 2 + (9)5 = 2 + 45 = 47$. The $10^{th}$ term of the AP is 47. ### 6. Triangles - **Similarity of Triangles:** Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). - **Criteria for Similarity:** - **AA (Angle-Angle) Similarity:** If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. - **SSS (Side-Side-Side) Similarity:** If the corresponding sides of two triangles are proportional, then they are similar. - **SAS (Side-Angle-Side) Similarity:** If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. - **Basic Proportionality Theorem (Thales Theorem):** If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. - **Converse of BPT:** If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. - **Areas of Similar Triangles Theorem:** The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. - If $\triangle ABC \sim \triangle DEF$, then $\frac{Area(\triangle ABC)}{Area(\triangle DEF)} = (\frac{AB}{DE})^2 = (\frac{BC}{EF})^2 = (\frac{CA}{FD})^2$. - **Pythagoras Theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides ($h^2 = p^2 + b^2$). - **Converse of Pythagoras Theorem:** If in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. **Example Question:** In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 1.5$ cm, $DB = 3$ cm, and $AE = 1$ cm, find $EC$. **Solution:** Since $DE \parallel BC$, by Basic Proportionality Theorem (Thales Theorem), we have: $\frac{AD}{DB} = \frac{AE}{EC}$ Substitute the given values: $\frac{1.5}{3} = \frac{1}{EC}$ $1.5 \times EC = 3 \times 1$ $EC = \frac{3}{1.5} = 2$ cm. ### 7. Coordinate Geometry - **Distance Formula:** The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ - **Section Formula:** The coordinates of a point $P(x,y)$ that divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1:m_2$ are: $$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1+m_2}$$ - **Mid-point Formula:** A special case of the section formula when $m_1=m_2=1$. The midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is: $$x = \frac{x_1+x_2}{2}, \quad y = \frac{y_1+y_2}{2}$$ - **Area of a Triangle:** The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: $$Area = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$ - If the area is 0, the three points are collinear. **Example Question:** Find the distance between the points $(2, 3)$ and $(4, 1)$. **Solution:** Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 1)$. Using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ $d = \sqrt{(4-2)^2 + (1-3)^2}$ $d = \sqrt{(2)^2 + (-2)^2}$ $d = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units. ### 8. Introduction to Trigonometry - **Trigonometric Ratios (for a right-angled triangle):** - $\sin A = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{P}{H}$ - $\cos A = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{B}{H}$ - $\tan A = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{P}{B}$ - $\csc A = \frac{1}{\sin A} = \frac{H}{P}$ - $\sec A = \frac{1}{\cos A} = \frac{H}{B}$ - $\cot A = \frac{1}{\tan A} = \frac{B}{P}$ - **Relations:** $\tan A = \frac{\sin A}{\cos A}$, $\cot A = \frac{\cos A}{\sin A}$. - **Trigonometric Identities:** 1. $\sin^2 A + \cos^2 A = 1$ 2. $1 + \tan^2 A = \sec^2 A$ 3. $1 + \cot^2 A = \csc^2 A$ - **Trigonometric Ratios of Complementary Angles:** - $\sin(90^\circ - A) = \cos A$ - $\cos(90^\circ - A) = \sin A$ - $\tan(90^\circ - A) = \cot A$ - $\cot(90^\circ - A) = \tan A$ - $\sec(90^\circ - A) = \csc A$ - $\csc(90^\circ - A) = \sec A$ - **Trigonometric Table (Common Angles):** | Angle $(\theta)$ | $0^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ | |:-----------------|:----------|:-----------|:-----------|:-----------|:-----------| | $\sin \theta$ | 0 | $1/2$ | $1/\sqrt{2}$ | $\sqrt{3}/2$ | 1 | | $\cos \theta$ | 1 | $\sqrt{3}/2$ | $1/\sqrt{2}$ | $1/2$ | 0 | | $\tan \theta$ | 0 | $1/\sqrt{3}$ | 1 | $\sqrt{3}$ | Undefined | **Example Question:** Evaluate: $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$. **Solution:** We know the values from the trigonometric table: $\sin 60^\circ = \sqrt{3}/2$ $\cos 30^\circ = \sqrt{3}/2$ $\sin 30^\circ = 1/2$ $\cos 60^\circ = 1/2$ Substitute these values: $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) + (\frac{1}{2})(\frac{1}{2})$ $= \frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1$. (This is also the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, so $\sin(60^\circ+30^\circ) = \sin 90^\circ = 1$). ### 9. Applications of Trigonometry (Heights and Distances) - **Angle of Elevation:** The angle formed by the line of sight with the horizontal when the object is above the horizontal level. - **Angle of Depression:** The angle formed by the line of sight with the horizontal when the object is below the horizontal level. - **Key Strategy:** 1. Draw a clear, labeled diagram representing the problem. 2. Identify the right-angled triangles formed. 3. Use appropriate trigonometric ratios ($\sin, \cos, \tan$) to relate the given angles and sides to the unknown quantities. 4. Solve the resulting equations. - **Important:** Angles of elevation and depression are always measured with respect to a horizontal line. **Example Question:** A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower. **Solution:** Let $AB$ be the tower of height $h$ meters. Let $C$ be the point on the ground 15 m away from the foot of the tower, so $BC = 15$ m. The angle of elevation $\angle ACB = 60^\circ$. In the right-angled triangle $ABC$: $\tan C = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$ $\tan 60^\circ = \frac{h}{15}$ We know $\tan 60^\circ = \sqrt{3}$. $\sqrt{3} = \frac{h}{15}$ $h = 15\sqrt{3}$ meters. The height of the tower is $15\sqrt{3}$ m. ### 10. Circles - **Tangent to a Circle:** A line that intersects the circle at exactly one point, called the point of contact. - **Secant to a Circle:** A line that intersects the circle at two distinct points. - **Theorems Related to Tangents:** 1. **Theorem 1:** The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. **Theorem 2:** The lengths of tangents drawn from an external point to a circle are equal. - **Properties derived from theorems:** - The center of the circle lies on the angle bisector of the angle between the two tangents from an external point. - If two tangents are drawn from an external point, the line segment joining the center to the external point bisects the angle between the tangents and the angle subtended by the chord of contact at the center. **Example Question:** A point P is 10 cm from the center of a circle. The length of the tangent drawn from P to the circle is 8 cm. Find the radius of the circle. **Solution:** Let O be the center of the circle and R be its radius. Let T be the point of contact of the tangent from P to the circle. According to Theorem 1, the radius is perpendicular to the tangent at the point of contact, so $\angle OTP = 90^\circ$. We have a right-angled triangle $\triangle OTP$. Given: $OP = 10$ cm (distance from center to point P), $PT = 8$ cm (length of tangent). By Pythagoras Theorem in $\triangle OTP$: $OT^2 + PT^2 = OP^2$ $R^2 + 8^2 = 10^2$ $R^2 + 64 = 100$ $R^2 = 100 - 64 = 36$ $R = \sqrt{36} = 6$ cm. The radius of the circle is 6 cm. ### 11. Areas Related to Circles - **Circumference of a Circle:** $C = 2\pi r$ - **Area of a Circle:** $A = \pi r^2$ - **Length of an Arc of a Sector:** $L = \frac{\theta}{360^\circ} \times 2\pi r$, where $\theta$ is the angle of the sector in degrees. - **Area of a Sector:** $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2$ - Alternatively, $A_{sector} = \frac{1}{2} \times L \times r$ (where L is arc length). - **Area of a Segment:** Area of minor segment = Area of the corresponding sector - Area of the corresponding triangle. - For acute angle $\theta$: $A_{segment} = \frac{\theta}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin\theta$ - For a semicircle, $\theta = 180^\circ$. For a quadrant, $\theta = 90^\circ$. **Example Question:** A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the minor segment. (Use $\pi = 3.14$) **Solution:** Given radius $r = 10$ cm and angle subtended at center $\theta = 90^\circ$. Area of sector $OAPB = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times 3.14 \times (10)^2$ $= \frac{1}{4} \times 3.14 \times 100 = 3.14 \times 25 = 78.5$ cm$^2$. Area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height}$. Since $\angle AOB = 90^\circ$, $OA$ and $OB$ are perpendicular. Area of $\triangle OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50$ cm$^2$. Area of minor segment = Area of sector $OAPB$ - Area of $\triangle OAB$ $= 78.5 - 50 = 28.5$ cm$^2$. ### 12. Surface Areas and Volumes - **Cuboid:** - Volume ($V$) = $l \times b \times h$ - Lateral Surface Area (LSA) / Area of 4 walls = $2h(l+b)$ - Total Surface Area (TSA) = $2(lb+bh+hl)$ - **Cube:** - Volume ($V$) = $a^3$ - Lateral Surface Area (LSA) = $4a^2$ - Total Surface Area (TSA) = $6a^2$ - **Right Circular Cylinder:** - Volume ($V$) = $\pi r^2 h$ - Curved Surface Area (CSA) = $2\pi r h$ - Total Surface Area (TSA) = $2\pi r(r+h)$ - **Right Circular Cone:** - Volume ($V$) = $\frac{1}{3}\pi r^2 h$ - Slant height ($l$) = $\sqrt{r^2+h^2}$ - Curved Surface Area (CSA) = $\pi r l$ - Total Surface Area (TSA) = $\pi r(l+r)$ - **Sphere:** - Volume ($V$) = $\frac{4}{3}\pi r^3$ - Surface Area (CSA = TSA) = $4\pi r^2$ - **Hemisphere:** - Volume ($V$) = $\frac{2}{3}\pi r^3$ - Curved Surface Area (CSA) = $2\pi r^2$ - Total Surface Area (TSA) = $3\pi r^2$ - **Frustum of a Cone:** (If $R$ and $r$ are radii of the two bases, $h$ is height, $l$ is slant height) - Volume ($V$) = $\frac{1}{3}\pi h (R^2 + r^2 + Rr)$ - Slant height ($l$) = $\sqrt{h^2 + (R-r)^2}$ - Curved Surface Area (CSA) = $\pi l (R+r)$ - Total Surface Area (TSA) = $\pi l (R+r) + \pi R^2 + \pi r^2$ - **Conversion of Solids:** When one solid is recast into another, its volume remains the same. **Example Question:** A solid metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. **Solution:** Volume of the sphere = $\frac{4}{3}\pi r_{sphere}^3$ $V_{sphere} = \frac{4}{3}\pi (4.2)^3$ cm$^3$. Volume of the cylinder = $\pi r_{cylinder}^2 h$ $V_{cylinder} = \pi (6)^2 h$ cm$^3$. Since the sphere is melted and recast into a cylinder, their volumes are equal: $V_{sphere} = V_{cylinder}$ $\frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h$ $\frac{4}{3} \times (4.2 \times 4.2 \times 4.2) = 36 \times h$ $4 \times (1.4 \times 4.2 \times 4.2) = 36 \times h$ $h = \frac{4 \times 1.4 \times 4.2 \times 4.2}{36}$ $h = \frac{1.4 \times 4.2 \times 4.2}{9}$ $h = 1.4 \times 0.466... \times 4.2 \approx 2.744$ cm. The height of the cylinder is approximately 2.74 cm. ### 13. Statistics - **Measures of Central Tendency:** Mean, Median, Mode. - **Mean ($\bar{x}$):** - **Ungrouped Data:** $\bar{x} = \frac{\sum x_i}{n}$ - **Grouped Data:** - **Direct Method:** $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ (where $x_i$ is class mark) - **Assumed Mean Method:** $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$. - **Step-deviation Method:** $\bar{x} = a + (\frac{\sum f_i u_i}{\sum f_i})h$, where $u_i = \frac{x_i - a}{h}$ and $h$ is the class size. - **Median:** The middle-most value of the data when arranged in ascending or descending order. - **Ungrouped Data:** If $n$ is odd, median is $(\frac{n+1}{2})^{th}$ term. If $n$ is even, median is average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ terms. - **Grouped Data:** $Median = L + (\frac{N/2 - cf}{f})h$ - $L$: lower limit of median class - $N$: total frequency ($\sum f_i$) - $cf$: cumulative frequency of the class preceding the median class - $f$: frequency of the median class - $h$: class size (assuming uniform class size) - **Mode:** The value that appears most frequently in the data. - **Ungrouped Data:** The observation with the highest frequency. - **Grouped Data:** $Mode = L + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2})h$ - $L$: lower limit of the modal class (class with highest frequency) - $f_1$: frequency of the modal class - $f_0$: frequency of the class preceding the modal class - $f_2$: frequency of the class succeeding the modal class - $h$: class size - **Empirical Relationship between the three measures:** $3 \times Median = Mode + 2 \times Mean$ - **Ogive (Cumulative Frequency Curve):** Graphical representation of cumulative frequency distribution. - **"Less than" Ogive:** Plot upper class limits against less than cumulative frequencies. - **"More than" Ogive:** Plot lower class limits against more than cumulative frequencies. - The median can be found graphically as the x-coordinate of the intersection point of the "less than" and "more than" ogives. **Example Question:** Find the mode of the following data: | Class Interval | Frequency | |:---------------|:----------| | 0-10 | 5 | | 10-20 | 8 | | 20-30 | 12 | | 30-40 | 7 | | 40-50 | 3 | **Solution:** The maximum frequency is 12, which corresponds to the class interval 20-30. So, the modal class is 20-30. $L = 20$ (lower limit of modal class) $h = 10$ (class size) $f_1 = 12$ (frequency of modal class) $f_0 = 8$ (frequency of class preceding the modal class) $f_2 = 7$ (frequency of class succeeding the modal class) Using the mode formula: $Mode = L + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2})h$ $Mode = 20 + (\frac{12 - 8}{2(12) - 8 - 7})10$ $Mode = 20 + (\frac{4}{24 - 15})10$ $Mode = 20 + (\frac{4}{9})10$ $Mode = 20 + \frac{40}{9} = 20 + 4.44... = 24.44$ (approx). The mode of the given data is approximately 24.44. ### 14. Probability - **Definition:** The probability of an event E, denoted as $P(E)$, is given by: $$P(E) = \frac{\text{Number of favorable outcomes to E}}{\text{Total number of possible outcomes}}$$ - **Range of Probability:** $0 \le P(E) \le 1$. - $P(E) = 0$ for an impossible event. - $P(E) = 1$ for a sure (certain) event. - **Complementary Events:** For any event E, $P(E) + P(\text{not } E) = 1$, or $P(\bar{E}) = 1 - P(E)$. - **Elementary Event:** An event having only one outcome. The sum of the probabilities of all elementary events of an experiment is 1. **Example Question:** A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red? **Solution:** Total number of balls in the bag = Number of red balls + Number of black balls = $3 + 5 = 8$. (i) Let E be the event that a red ball is drawn. Number of favorable outcomes (red balls) = 3. $P(\text{E}) = P(\text{red ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8}$. (ii) Let $\bar{E}$ be the event that the ball drawn is not red. $P(\text{not red}) = 1 - P(\text{red})$ $P(\text{not red}) = 1 - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8}$. Alternatively, number of not red balls (black balls) = 5. $P(\text{not red}) = \frac{5}{8}$.