24. Probability Previous Year Questions

For any event E, 0 ≤ P(E) ≤ 1. Here 5/4 = 1.25 > 1, which cannot be a probability. [1 Mark]

Total outcomes n(S) = 30.
(a) Perfect squares {1, 4, 9, 16, 25} (5 numbers): P = 5/30 = 1/6. [1 Mark]
(b) Even multiples of 3 {6, 12, 18, 24, 30} (5 numbers): P = 5/30 = 1/6. [1 Mark]
(c) Prime numbers {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} (10 numbers): P = 10/30 = 1/3. [1 Mark]

Correct Option: (b) A is false, R is true.
P(even) = 3/6 = 1/2 ≠ 2/3 (Assertion false). Even numbers = {2, 4, 6} (Reason true). [1 Mark]

Total letters n(S) = 8.
(a) Letters {A, D, M, N, O, Y} ⇒ 6/8 = ³/₄. [1 Mark]
(b) Letters {S, U} ⇒ 2/8 = ¹/₂. [1 Mark]
(c) Letters in both SUNDAY and MONDAY = {A, D, N, Y} ⇒ 4/8 = ¹/₂. [1 Mark]

Total number of letters in English alphabet $n(S) = 26$.
Vowels are $\{A, E, I, O, U\}$, so favorable outcomes $n(E) = 5$.
Probability $P(E) = \frac{n(E)}{n(S)} = \frac{5}{26}$. Correct Option: (b) [1 Mark]

Sample space $S = \{1, 2, 3, \dots, 25\} \implies n(S) = 25$.

(a) Multiple of 5 $E_1 = \{5, 10, 15, 20, 25\} \implies n(E_1) = 5$.
$P(E_1) = \frac{5}{25} = \mathbf{\frac{1}{5}}$.1 Mark

(b) Perfect Square $E_2 = \{1, 4, 9, 16, 25\} \implies n(E_2) = 5$.
$P(E_2) = \frac{5}{25} = \mathbf{\frac{1}{5}}$.1 Mark

(c) Prime Number $E_3 = \{2, 3, 5, 7, 11, 13, 17, 19, 23\} \implies n(E_3) = 9$.
$P(E_3) = \mathbf{\frac{9}{25}}$.1 Mark

Total letters in ‘AUTHORIZES’ $= 10$. Total outcomes $n(S) = 10$.
(i) Vowels in ‘AUTHORIZES’ are: A, U, O, I, E $\implies n(E_1) = 5$.
$P(\text{vowel}) = \frac{5}{10} = \mathbf{\frac{1}{2}}$ (or $0.5$). [1 Mark]
(ii) First 9 letters of English alphabet: {A, B, C, D, E, F, G, H, I}.
Letters in word from this set: A, H, I, E $\implies n(E_2) = 4$.
$P = \frac{4}{10} = \mathbf{\frac{2}{5}}$ (or $0.4$). [1 Mark]
(iii) Last 9 letters of English alphabet (18th to 26th): {R, S, T, U, V, W, X, Y, Z}.
Letters in word from this set: U, T, R, Z, S $\implies n(E_3) = 5$.
$P = \frac{5}{10} = \mathbf{\frac{1}{2}}$ (or $0.5$). [1 Mark]

Step 1 Sample space $S = \{1, 2, 3, \dots, 25\} \implies n(S) = 25$.
(i) Event $A$ (odd numbers) = $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\} \implies n(A) = 13$.
$\mathbf{P(A) = \frac{13}{25}}$. [1 Mark]
Step 2 (ii) Event $B$ (divisible by 2 and 3 both $\implies$ divisible by $\text{LCM}(2, 3) = 6$):
$B = \{6, 12, 18, 24\} \implies n(B) = 4$.
$\mathbf{P(B) = \frac{4}{25}}$. [1 Mark]
Step 3 (iii) Event $C$ (numbers less than 16):
$C = \{1, 2, 3, \dots, 15\} \implies n(C) = 15$.
$\mathbf{P(C) = \frac{15}{25} = \frac{3}{5}}$. [1 Mark]

Solution The Sun rising in the west is an impossible physical event, and since $P(\phi) = 0$, (R) directly explains (A).
Correct Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 Mark]

Step 1 Sample Space:
$S = \{\text{'B'}, \text{'I'}, \text{'N'}, \text{'O'}, \text{'C'}, \text{'U'}, \text{'L'}, \text{'A'}, \text{'R'}, \text{'S'}\} \implies n(S) = 10$. [1 Mark]
Step 2 (a) Event $E_1$ (vowels in the word):
$E_1 = \{\text{'I'}, \text{'O'}, \text{'U'}, \text{'A'}\} \implies n(E_1) = 4$.
$\mathbf{P(E_1) = \frac{4}{10} = \frac{2}{5}}$. [1 Mark]
Step 3 (b) Event $E_2$ (first 8 letters of English alphabet $\{A, B, C, D, E, F, G, H\}$ present in word):
$E_2 = \{\text{'B'}, \text{'C'}, \text{'A'}\} \implies n(E_2) = 3$.
$\mathbf{P(E_2) = \frac{3}{10}}$. [1 Mark]

Sample Space $S = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} \implies n(S) = 10$.
(i) Prime number: The only prime in the set is $2$.
Favourable outcomes: $\{2\} \implies n(E_1) = 1$.
$$P(\text{prime}) = \frac{n(E_1)}{n(S)} = \mathbf{\frac{1}{10}}$$ [1 Mark]
(ii) Number divisible by 4:
Favourable outcomes: $\{4, 8, 12, 16, 20\} \implies n(E_2) = 5$.
$$P(\text{divisible by 4}) = \frac{5}{10} = \mathbf{\frac{1}{2}}$$ [1 Mark]
(iii) Number that is a multiple of 6:
Favourable outcomes: $\{6, 12, 18\} \implies n(E_3) = 3$.
$$P(\text{multiple of 6}) = \mathbf{\frac{3}{10}}$$ [1 Mark]
(iv) An odd number:
All numbers in the bag are even $\{2, 4, \dots, 20\}$, so there are no odd numbers.
Favourable outcomes: $\emptyset \implies n(E_4) = 0$.
$$P(\text{odd number}) = \frac{0}{10} = \mathbf{0}$$ [1 Mark]
Answers: (i) $\frac{1}{10}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{10}$, (iv) $0$.

Sample Space $S = \{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p\}$, $n(S) = 16$.
(i) Vowel Vowels in $S$: $\{a, e, i, o\} \implies 4$ outcomes.
$$P(\text{vowel}) = \frac{4}{16} = \mathbf{\frac{1}{4}}$$ [1 Mark]
(ii) Consonant Consonants in $S$: $16 - 4 = 12$ outcomes.
$$P(\text{consonant}) = \frac{12}{16} = \mathbf{\frac{3}{4}}$$ [1 Mark]
(iii) None of letters in 'median' Letters $\{m, e, d, i, a, n\} = 6$ cards.
Cards containing none of these letters $= 16 - 6 = 10$.
$$P(\text{none of the letters of median}) = \frac{10}{16} = \mathbf{\frac{5}{8}}$$ [1 Mark]
Answers: (i) $\frac{1}{4}$, (ii) $\frac{3}{4}$, (iii) $\frac{5}{8}$.