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A die is thrown, find the probability of following events:

Question:

A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

Solution:

The sample space of the given experiment is given by

S = {1, 2, 3, 4, 5, 6}

(i) Let A be the event of the occurrence of a prime number.

Accordingly, A = {2, 3, 5}

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{6}=\frac{1}{2}$

(ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}

$\therefore \mathrm{P}(\mathrm{B})=\frac{\text { Number of outcomes favourable to } \mathrm{B}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{4}{6}=\frac{2}{3}$

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C = {1}

$\therefore \mathrm{P}(\mathrm{C})=\frac{\text { Number of outcomes favourable to } \mathrm{C}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{1}{6}$

(iv) Let D be the event of the occurrence of a number greater than 6.

Accordingly, $D=\Phi$

$\therefore \mathrm{P}(\mathrm{D})=\frac{\text { Number of outcomes favourable to } \mathrm{D}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{D})}{n(\mathrm{~S})}=\frac{0}{6}=0$

(v) Let E be the event of the occurrence of a number less than 6.

Accordingly, E = {1, 2, 3, 4, 5}

$\therefore \mathrm{P}(\mathrm{E})=\frac{\text { Number of outcomes favourable to } \mathrm{E}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{5}{6}$

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