Data available for profit (₹ Thousands) of a company as | Year | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | |---|---|---|---|---|---|---|---| | Profit(₹ 000) | 114 | 130 | 126 | 144 | 138 | 156 | 164 | Based on the above data using least square method the trend value for the year '2007' is

CUET UG 2025 — Mathematics Statistics & Applications

2025mcqmedium

Data available for profit (₹ Thousands) of a company as

Year	2004	2005	2006	2007	2008	2009	2010
Profit(₹ 000)	114	130	126	144	138	156	164

Based on the above data using least square method the trend value for the year '2007' is

Official previous-year question

Held on 2 Jun 2025 · Verified 13 Jul 2026.

Options

A
147.64
B
138.86
C
130.57
D
131.22

Did you get this right?

Solution

Statement (A): $P(S | B) = 1$

Using the conditional probability formula:

$P(S | B) = \frac{P(S \cap B)}{P(B)}$

Since $B \subseteq S$ , we have $S \cap B = B$

$P(S | B) = \frac{P(B)}{P(B)} = 1$

Statement (A) is TRUE.

Statement (B): $P(A \cap B) = P(A) + P(B) + P(A \cup B)$

The addition rule for probability states:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

The given statement has a plus sign instead of a minus sign.

Statement (B) is FALSE.

Statement (C): $P(\bar{A} | B) = 1 - P(A | B)$

Given that event $B$ has occurred, either $A$ occurs or $\bar{A}$ occurs:

$P(A | B) + P(\bar{A} | B) = 1$

Therefore:

$P(\bar{A} | B) = 1 - P(A | B)$

Statement (C) is TRUE.

Statement (D): $P(A | B) = \frac{P(A \cap B)}{P(B)}, P(B) \neq 0$

This is the definition of conditional probability.

Statement (D) is TRUE.

Statements (A), (C), and (D) are TRUE.

Statement (B) is FALSE.

The correct answer is: (A), (C) and (D) only.

Your note

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Work through every CUET UG Statistics & Applications PYQ, year by year.

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Year

2004

2005

2006

2007

2008

2009

2010

Profit(₹ 000)

114

130

126

144

138

156

164

Solution

Statement (A): $P(S | B) = 1$

Using the conditional probability formula:

$P(S | B) = \frac{P(S \cap B)}{P(B)}$

Since $B \subseteq S$ , we have $S \cap B = B$

$P(S | B) = \frac{P(B)}{P(B)} = 1$

Statement (A) is TRUE.

Statement (B): $P(A \cap B) = P(A) + P(B) + P(A \cup B)$

The addition rule for probability states:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

The given statement has a plus sign instead of a minus sign.

Statement (B) is FALSE.

Statement (C): $P(\bar{A} | B) = 1 - P(A | B)$

Given that event $B$ has occurred, either $A$ occurs or $\bar{A}$ occurs:

$P(A | B) + P(\bar{A} | B) = 1$

Therefore:

$P(\bar{A} | B) = 1 - P(A | B)$

Statement (C) is TRUE.

Statement (D): $P(A | B) = \frac{P(A \cap B)}{P(B)}, P(B) \neq 0$

This is the definition of conditional probability.

Statement (D) is TRUE.

Statements (A), (C), and (D) are TRUE.

Statement (B) is FALSE.

The correct answer is: (A), (C) and (D) only.

CUET UG 2025 — Mathematics Statistics & Applications

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Solution

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CUET UG 2025 — Mathematics Statistics & Applications

Options

Did you get this right?

Solution

Your note

CUET UG Statistics & Applications in other years

More CUET UG Statistics & Applications questions

Other Mathematics topics for CUET UG

Algebra

Calculus

Applied-Mathematics

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Statistics & Applications