Find the probability that a number selected at random.

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Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a
(i) prime number
(ii) multiple of 7
(iii) multiple of 3 or 5.

Solution:

Numbers are 1, 2, 3, 4, 5,…..30, 31, 32, 33, 34, 35
Total = 35
(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
which are 11
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 11 }{ 35 }
(ii) Multiple of 7 are 7, 14, 21, 28, 35 which are 5
Probability of multiple of 7 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 5 }{ 35 } = \\ \frac { 1 }{ 7 }
(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12 ,15, 18, 20, 21, 24, 25, 27, 30, 33, 35.
Which are 16 in numbers
Probability of multiple of 3 or 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 16 }{ 35 }

Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is
(i) divisible by 5
(ii) a number which is a perfect square.

Solution:

Number of cards which are marked with numbers
13, 14, 15, 16, 17,….to 59, 60 are = 48
(i) Numbers which are divisible by 5 will be
15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
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 { 10 }{ 48 } = \\ \frac { 5 }{ 24 }
(ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers.
Probability of number which is a perfect square will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 4 }{ 48 } = \\ \frac { 1 }{ 12 }

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