Question

A die is thrown, find the probability of the 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.

Answer 1

A die is thrown, find the probability of the 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.

The sample space of the given experiment is given by
$S = \\{1, 2, 3, 4, 5, 6\\}$

(i) Let A$$$ be the event of the occurrence of a prime number. Accordingly, A = \{2, 3, 5\}$∴$∴P(A)=\dfrac{\text{Number } \text{of }\text{ outcomes }\text{ favorable} \text{ to }A}{\text{Total } \text{number }\text{ of}\text{ possible }\text{outcomes }}$$=\dfrac{n(A)}{n(S)}$

$=\dfrac{3}{6}$

$=\dfrac{1}{2}$

(ii) Let $B$ be the event of the occurrence of a number greater than or equal to $3.$ Accordingly, $B = \\{3, 4, 5, 6\\}$
$∴P(B)=\dfrac{\text{Number } \text{of }\text{ outcomes }\text{ favorable} \text{ to }B}{\text{Total } \text{number }\text{ of}\text{ possible }\text{outcomes }}$
$=\dfrac{n(A)}{n(S)}$

$=\dfrac{4}{6}$

$=\dfrac{2}{3}$

(iii) Let $C$ be the event of the occurrence of a number less than or equal to one. Accordingly, $C = \\{1\\}$
$∴P(C)=\dfrac{\text{Number } \text{of }\text{ outcomes }\text{ favorable} \text{ to }C}{\text{Total } \text{number }\text{ of}\text{ possible }\text{outcomes }}$
$=\dfrac{n(C)}{n(S)}$
$=\dfrac{1}{6}$
(iv) Let $E = \\{1, 2, 3, 4, 5\\}$D be the event of the occurrence of a number greater than 6. Accordingly, $D = Φ$
$∴P(D)=\dfrac{\text{Number } \text{of }\text{ outcomes }\text{ favorable} \text{ to }D}{\text{Total } \text{number }\text{ of}\text{ possible }\text{outcomes }}$
$=\dfrac{n(D)}{n(S)}$

$=\dfrac{0}{6}$

$=0$

(v) Let $E$ be the event of the occurrence of a number less than 6. Accordingly, $E = \\{1, 2, 3, 4, 5\\}$
$∴P(E)=\dfrac{\text{Number } \text{of }\text{ outcomes }\text{ favorable} \text{ to }E}{\text{Total } \text{number }\text{ of}\text{ possible }\text{outcomes }}$

$=\dfrac{n(E)}{n(S)}$

$=\dfrac{5}{6}$