Probability — NCERT Solutions, AI Tutor & Practice

By the basic properties of probability:

(i) A certain (sure) event always happens, so its probability is $1$.

(ii) An impossible event never happens, so its probability is $0$.

(iii) The elementary events cover all possible outcomes, so their probabilities add up to $1$.

(iv) For any event $E$, $0 \le P(E) \le 1$.

When a fair coin is tossed once, the sample space is $\{H, T\}$, so the total number of equally likely outcomes is $2$.

(i) Favourable outcomes for a head $= 1$. $P(\text{head}) = \frac{1}{2}$

(ii) Favourable outcomes for a tail $= 1$. $P(\text{tail}) = \frac{1}{2}$

A die has outcomes $\{1, 2, 3, 4, 5, 6\}$, so total outcomes $= 6$.

(i) Prime numbers on a die are $2, 3, 5$ — that is $3$ outcomes. $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$

(ii) Numbers lying strictly between 2 and 6 are $3, 4, 5$ — that is $3$ outcomes. $P(\text{between 2 and 6}) = \frac{3}{6} = \frac{1}{2}$

(iii) Odd numbers are $1, 3, 5$ — that is $3$ outcomes. $P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$

Total number of cards $= 52$.

(i) There are $4$ kings. $P(\text{king}) = \frac{4}{52} = \frac{1}{13}$

(ii) Red face cards are the King, Queen and Jack of hearts and diamonds $= 2 \times 3 = 6$ cards. $P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$

(iii) Spades $= 13$. Aces not already counted in spades $= 3$ (hearts, diamonds, clubs). So cards that are a spade OR an ace $= 13 + 3 = 16$. Cards that are neither $= 52 - 16 = 36$. $P(\text{neither spade nor ace}) = \frac{36}{52} = \frac{9}{13}$

Total balls $= 5 + 8 + 7 = 20$.

(i) White balls $= 8$. $P(\text{white}) = \frac{8}{20} = \frac{2}{5}$

(ii) $P(\text{green}) = \dfrac{7}{20}$, so $P(\text{not green}) = 1 - \frac{7}{20} = \frac{13}{20}$

(iii) Red or white $= 5 + 8 = 13$ balls. $P(\text{red or white}) = \frac{13}{20}$

Since $E$ and 'not $E$' are complementary events: $P(\text{not } E) = 1 - P(E) = 1 - 0.32 = 0.68$

No, a probability cannot be $1.4$. For any event, $0 \le P(E) \le 1$. Since $1.4 > 1$, it is not a valid probability.

When two dice are thrown, the total number of equally likely outcomes $= 6 \times 6 = 36$.

(i) Sum $= 7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — $6$ outcomes. $P(\text{sum }7) = \frac{6}{36} = \frac{1}{6}$

(ii) Doublets: $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ outcomes. $P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}$

(iii) Sum greater than 10 means sum is 11 or 12: $(5,6),(6,5),(6,6)$ — $3$ outcomes. $P(\text{sum} > 10) = \frac{3}{36} = \frac{1}{12}$

Numbers from 3 to 27 inclusive: total $= 27 - 3 + 1 = 25$ cards.

(i) Even numbers from 3 to 27: $4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26$ — that is $12$ numbers. $P(\text{even}) = \frac{12}{25}$

(ii) Prime numbers less than 20 within this range: $3, 5, 7, 11, 13, 17, 19$ — that is $7$ numbers. $P(\text{prime} < 20) = \frac{7}{25}$

For geometric probability, probability $= \dfrac{\text{favourable area}}{\text{total area}}$.

Area of larger circle $= \pi (21)^2 = \pi \times 441$.

Area of smaller circle $= \pi (7)^2 = \pi \times 49$.

$P(\text{inside small circle}) = \frac{\pi \times 49}{\pi \times 441} = \frac{49}{441} = \frac{1}{9}$

(i) Defective bulbs $= 4$, total $= 20$. $P(\text{defective}) = \frac{4}{20} = \frac{1}{5}$

(ii) After drawing one defective bulb and not replacing it:

• Remaining bulbs $= 20 - 1 = 19$
• Non-defective bulbs $= 20 - 4 = 16$ (unchanged, since the removed bulb was defective)

$P(\text{second not defective}) = \frac{16}{19}$

Total faces $= 6$.

(i) The letter A appears on $2$ faces. $P(A) = \frac{2}{6} = \frac{1}{3}$

(ii) The letter D appears on $1$ face. $P(D) = \frac{1}{6}$

Total cards $= 100$.

(i) Perfect squares from 1 to 100: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ — that is $10$ numbers. $P(\text{perfect square}) = \frac{10}{100} = \frac{1}{10}$

(ii) Two-digit numbers divisible by 5 are $10, 15, 20, \ldots, 95$. These form an AP with first term $10$, common difference $5$, last term $95$. Number of terms $= \dfrac{95 - 10}{5} + 1 = \dfrac{85}{5} + 1 = 17 + 1 = 18$. $P(\text{two-digit multiple of 5}) = \frac{18}{100} = \frac{9}{50}$