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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 6 · Set 3 · Q24
A bag contains balls numbered 2 to 91 such that each ball bears a different number. A ball is drawn at random from the bag. Find the probability that…
A bag contains balls numbered $\displaystyle 2$ to $\displaystyle 91$ such that each ball bears a different number. A ball is drawn at random from the bag. Find the probability that
(i)
it bears a $\displaystyle 2$- digit number
(ii)
it bears a multiple of 1.
Marking-scheme solution
Total possible outcomes = $\displaystyle 90$
(i) Number of favourable outcomes for a $\displaystyle 2$-digit number = $\displaystyle 82$
\[\mathrm{P} \text { (2-digit number) }=\frac{82}{90} \text { or } \frac{41}{45}
\]
(ii)
Number of favourable outcomes for multiple of $\displaystyle 1$ = $\displaystyle 90$
\(\displaystyle \mathrm{P}(\) a number multiple of $\displaystyle 1$\(\displaystyle )=\frac{90}{90}\) or $\displaystyle 1$
Adding equations we get
\[x+y=3
\]
Subtracting equations we get
\[-x+y=-1
\]
Solving to get
\(\displaystyle x=2\) and \(\displaystyle y=1\)
OR
Let smaller angle be \(\displaystyle x\) and greater angle be y
\[\text { ATQ, } x+y=180
\]
Also \(\displaystyle y=x+50\)
Solving we get
\[x=65^{\circ} \text { and } \mathrm{y}=115^{\circ}
\]
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.