SSC CGL 2025 Tier-I · 14 September 2025, Shift 2 · General Intelligence and Reasoning · Question 14 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

A box has 5 red, 3 blue, 2 green balls. Probability of drawing a non-green ball?

(a)0.5
(b)0.8
(c)0.6
(d)0.7

Answer

Why

Correct — B. Count the balls the question wants, then divide by all the balls.

Rule: probability = favourable outcomes ÷ total outcomes.

Total balls: 5 + 3 + 2 = 10
Non-green balls: 5 red + 3 blue = 8
P(non-green) = 8 ÷ 10 = 0.8

Check by complement: 1 − 2⁄10 = 1 − 0.2 = 0.8. That is option (b).

Why the others are wrong

(a)
0.5 — 0.5 is 5 ÷ 10, the chance of a red ball. Non-green also takes in the 3 blue balls, so the count is 8, not 5.
(c)
0.6 — 0.6 means 6 favourable balls out of 10. No group of these colours makes 6: the possible counts are 2, 3, 5, 7, 8 and 10.
(d)
0.7 — 0.7 is 7 ÷ 10, which is 1 − 3⁄10: the chance of not drawing blue. The question leaves out the 2 green balls, not the 3 blue ones.

Concept

Probability compares the outcomes you want with all the equally likely outcomes. Each ball is one outcome, so the total here is 10.

Non-green is a complement: every ball except the green ones. Count it directly (red + blue = 8) or subtract the green share from 1 (1 − 0.2). The two routes must agree, so the second is a free check on the first.

The stem does not say 'at random' or 'one ball'. The key takes the standard reading: one ball is drawn, and every ball is equally likely.

Key facts

Probability = favourable outcomes ÷ total outcomes.
P(not A) = 1 − P(A).
With 2 green balls among 10, P(green) = 0.2 and P(non-green) = 0.8.

Study next

Two balls drawn without replacement, where the second draw has one ball fewer'At least one' questions solved as 1 − P(none)Two-dice sums, where the total is 6 × 6 = 36 outcomes

Common traps

Leaving out the wrong colour: removing the 3 blue balls instead of the 2 green gives 0.7.
Counting the red balls alone as non-green, which gives 0.5.

Here a one-draw probability is set in the Reasoning section, stated in a single line, with four answer choices one tenth apart.

Related PYQs

No directly related past PYQ was found.