Chapter 4 · Kerala SSLC Class 10 Maths
Mathematics of Chance
Two ways to learn — a crisp formula reference, or a story that builds probability intuition from scratch.
Four scenes that build each idea from a real situation, step by step.
How this works
Each scene turns a chance situation into one clean rule. Read the story, then tap "The Maths" to see the formal result.
Two Boxes of Beads
A game stall has two boxes. Box A has 6 black beads and 5 white beads. Box B has 5 black beads and 4 white beads. You win if you draw a black bead. You get to pick which box to draw from.
Box A has more black beads than Box B — 6 versus 5. Box A must be the better choice.
Is that really true? Box A also has more beads overall. What actually matters isn't the raw count of black beads — it's what fraction of each box is black.
Box A: 6 black out of 11 total → P(black) = 6/11 ≈ 0.545
Box B: 5 black out of 9 total → P(black) = 5/9 ≈ 0.556
5/9 is actually slightly larger than 6/11. Box B — the one with FEWER black beads — is actually the better choice, because it has a larger fraction of black beads overall.
The Spinning Disc
A circular disc, divided into 8 equal sectors, is mounted so it can spin freely. 3 of the sectors are painted yellow. When the disc stops spinning, an arrow points to one sector at random.
That's easy — 3 out of 8 sectors are yellow, so P(yellow) = 3/8.
Right, because the sectors are all equal-sized regions and there's a clear count: 8 total, 3 favourable. But now imagine a different kind of randomness — not a countable set of sectors, but a continuous region.
Take a rectangular piece of cardboard. Draw a triangle by joining the midpoint of one side to the two corners of the opposite side. Close your eyes and mark a random point anywhere inside the rectangle.
There's no way to count points — there are infinitely many of them inside that rectangle!
Exactly right. When outcomes are points in a region rather than a countable list, probability has to be measured with area instead of counting.
The triangle shares the same base and height as the rectangle.
Area of triangle = (1/2) × base × height = half the rectangle's area.
P(point lands in triangle) = triangle area / rectangle area = 1/2
Two Slips, Two Boxes
Box A contains 4 slips, numbered 1 to 4. Box B contains 2 slips, numbered 1 and 2. One slip is drawn from each box.
How many different pairs of numbers could I possibly get?
Think of it position by position. If the first slip is 1, the second slip could be 1 or 2 — that's 2 possible pairs already, just from the first slip being 1. Now do the same for 2, 3, and 4 from Box A:
(1,1) (1,2)
(2,1) (2,2)
(3,1) (3,2)
(4,1) (4,2)
Total = 4 × 2 = 8 pairs
Each of the 4 choices from Box A pairs with each of the 2 choices from Box B — that's a multiplication, not an addition.
In how many of these 8 pairs are BOTH numbers odd?
Odd numbers in Box A: 1 and 3 (that's 2 of them). Odd numbers in Box B: just 1 (that's 1 of them). So the matching pairs are (1,1) and (3,1) — exactly 2 pairs.
Matching pairs = (odd count in A) × (odd count in B) = 2 × 1 = 2
P(both odd) = 2 / 8 = 1/4
The Mango Baskets
A shopkeeper has two baskets of mangoes. Basket A has 50 mangoes, 20 of which are unripe. Basket B has 40 mangoes, 15 of which are unripe. A customer picks one mango from each basket, without looking.
What's the probability that AT LEAST ONE of my two mangoes is ripe?
"At least one ripe" could happen in three different ways: ripe-and-unripe, unripe-and-ripe, or ripe-and-ripe. Counting all three separately is doable, but slow and easy to get wrong.
Easier question: in how many ways are BOTH mangoes unripe?
Basket A has 20 unripe out of 50; Basket B has 15 unripe out of 40. This is just the paired-events count from before:
Total pairs = 50 × 40 = 2000
Both unripe = 20 × 15 = 300
P(both unripe) = 300 / 2000
Now here's the trick: "at least one ripe" is the exact opposite of "both unripe". Between them, they cover every possibility.
P(at least one ripe) = 1 − P(both unripe)
= 1 − 300/2000
= 1700/2000 = 17/20
The Big Picture
Every idea here branches from one formula: favourable ÷ total. Comparing boxes, measuring regions with area, combining two draws, and answering "at least one" questions are all just different situations feeding into that same ratio.
Compare two boxes?
Convert both to fractions first, then compare
Outcome is a region, not a count?
P(E) = favourable area / total area
Drawing from two boxes at once?
total pairs = n₁ × n₂
"At least one" of two draws?
1 − P(neither)
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