Determine the probability p,

Solution:

(a) When a fair die is thrown, the possible outcomes are $S=\{1,2,3,4,5,6\}$

$\therefore$ total outcomes $=6$ and the odd numbers are $1,3,5$

$\therefore$ Favourable outcomes $=3$

We know that,

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required probability $=\frac{3}{6}=\frac{1}{2}$

(b) When a fair coin is tossed two times, the sample space is $\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$

$\therefore$ Total outcomes $=4$

If at least one head appears then the favourable cases are $\mathrm{HH}, \mathrm{HT}$ and $\mathrm{TH}$.

$\therefore$ Favourable outcomes $=3$

We know that,

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required probability $=\frac{3}{4}$

(c) When a pair of dice is rolled, total number of cases

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Total Sample Space, n(S) = 36

If sum is 6 then possible outcomes are (1,5), (2,4), (3,3), (4,2) and (5,1).

∴ Favourable outcomes = 5

We know that,

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required probability $=\frac{5}{36}$

objective TYPE questions: