Find the probability that the number on the ticket is:

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Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.

Solution:

In a box there are 14 tickets with number
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Number of possible outcomes = 14
(i) Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 9 in number
Probability of prime will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 9 }{ 14 }
(ii) Number less than 16 are 3, 5, 7, 9, 11, 13, 15
which are 7 in numbers,
Probability of number less than 16 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 7 }{ 14 }
= \\ \frac { 1 }{ 2 }
(iii) Numbers divisible by 3 are 3, 9, 15, 21, 27
which are 5 in number
Probability of number divisible by 3 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 5 }{ 14 }

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.

Solution:

There are 90 discs in a box containing numbered from 1 to 90.
Number of possible outcomes = 90
(i) Two digit numbers are 10 to 90 which are 81 in numbers.
Probability of two digit number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 81 }{ 90 }
= \\ \frac { 9 }{ 10 }
(ii) Perfect squares are 1, 4, 9, 16, 25, 36,49, 64, 81
which are 9 in numbers.
Probability of square will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 9 }{ 90 }
= \\ \frac { 1 }{ 10 }
(iii) Number divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
which are 18 in numbers.
Probability of number divisible by 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 18 }{ 90 }
= \\ \frac { 1 }{ 5 }

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