Question

Identical cards are marked 1 to 100. When a card is drawn at random what is the probability that it is
a. A multiple of 11.
b. A perfect square .
c. A number in which one digit is twice the other
d. A multiple of 2 and 3.
e. A prime number containing the digit 9​

Answer 1

Answer :-

$\tt{i.) \: \: \: \dfrac{9}{100} } \\ \\ \tt{ii.) \: \: \: \dfrac{1}{10} } \\ \\ \tt{iii.) \: \: \: \dfrac{2}{25} } \\ \\ \tt{iv.) \: \: \: \dfrac{1}{25} } \\ \\ \tt{v.) \: \: \: \dfrac{3}{50} }$

Formula to be used :-

$\tt{Probability= \dfrac{Number \: \: of \: \: favourable \: \: outcomes }{Total \: \: number \: \: of \: \: possible \: \: outcomes }}$

Solution :-

Total number of cards = 100

Total number of possible outcomes = Total number of cards = 100

i.) A multiple of 11

Multiples of 11 from 1 to 100 are 11, 22, 33, 44, 55, 66, 77, 88, and 99 (Total 9 in number).

Number of favourable outcomes = Multiples of 11 from 1 to 100 = 9

$= > \tt{Probability(a \: \: multiple \: \: of \: \: 11) = \dfrac{9}{100} }$

ii.) A perfect square

A perfect square from 1 to 100 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 (Total 10 in number).

Number of favourable outcomes = A perfect square from 1 to 100 = 10

$= > \tt{Probability(a \: \: perfect \: \: square \: \: 11) = \dfrac{10}{100} } = \dfrac{1}{10}$

iii.) A number in which one digit is twice the other.

The pairs formed by the numbers which satisfies the condition are (1, 2), (2, 4), (3, 6), (4, 8).

So, the numbers are - 12, 21, 24, 42, 36, 63, 48, 84 (Total 8 in number).

Number of favourable outcomes = A number in which one digit is twice the other from 1 to 100 = 8

$= > \tt{Probability(A \: \: number \: \: in \: \: which \: \: one \: \: digit \: \: is \: \: twice \: \: the \: \: other ) = \dfrac{8}{100} } = \dfrac{2}{25}$

iv.) A multiple of 2 and 3.

A multiple of 2 and 3 from 1 to 100 are 6, 12, 18, and 24 (Total 4 in number).

Number of favourable outcomes = A perfect square from 1 to 100 = 4

$= > \tt{Probability(A \: \: multiple \: \: of\: \: 2 \: \: and \: \: 3 ) = \dfrac{4}{100} } = \dfrac{1}{25}$

v.) A prime number containing the digit 9

A prime number containing the digit 9 from 1 to 100 are 19, 29, 59, 79, 89 and 97 (Total 6 in number).

Number of favourable outcomes = A prime number containing the digit 9 from 1 to 100 = 6

$= > \tt{Probability(A \: \: prime \: \: number\: \: containing\: \: 9) = \dfrac{6}{100} } = \dfrac{3}{50}$