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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 4 · Set 1 · Q27
Three coins are tossed simultaneously. What is the probability of getting(i)at least one head ?(ii)exactly two tails ?(iii)at most one tail ?A box contains $\displaystyle 90$ discs which are numbered $\displaystyle 1$ to $\displaystyle 90$ . If one disc is drawn at random from the box, find the probability that it bears a :(i)$\displaystyle 2$-digit number less than 40.(ii)number divisible by $\displaystyle 5$ and greater than 50.(iii)a perfect square number.
Three coins are tossed simultaneously. What is the probability of getting
(i)
at least one head ?
(ii)
exactly two tails ?
(iii)
at most one tail ?
A box contains $\displaystyle 90$ discs which are numbered $\displaystyle 1$ to $\displaystyle 90$ . If one disc is drawn at random from the box, find the probability that it bears a :
(i)
$\displaystyle 2$-digit number less than 40.
(ii)
number divisible by $\displaystyle 5$ and greater than 50.
(iii)
a perfect square number.
Marking-scheme solution
Total number of outcomes \(\displaystyle =8\)
(i) P (at least one head) \(\displaystyle =\frac{7}{8}\)
(ii) P (exactly $\displaystyle 2$ tails) \(\displaystyle =\frac{3}{8}\)
(iii) P (at most one tail) \(\displaystyle =\frac{4}{8}\) or \(\displaystyle \frac{1}{2}\)
Total outcomes = $\displaystyle 90$
(i) P ($\displaystyle 2$ digit number less than $\displaystyle 40$\(\displaystyle )=\frac{30}{90}\) or \(\displaystyle \frac{1}{3}\)
(ii) P (a number divisible by $\displaystyle 5$ and greater than $\displaystyle 50$\(\displaystyle )=\frac{8}{90}\) or \(\displaystyle \frac{4}{45}\)
(iii) P (a perfect square number) \(\displaystyle =\frac{9}{90}\) or \(\displaystyle \frac{1}{10}\)
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.