Question

a box contain 1 to 100 number cards if one card is drawn at random find the probability of the card will be
a perfect square
a prime number
a two digit number
a multiple of 9
​

Answer 1

let numbers on cards in box be sample space

${n(S)\:=\:1,2,3,4,5,6,7...,100}$

1. a perfect square : €vent A

p.er fect square are = 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100

${n(A)\:=\:9}$

${p(A)\:=\:\frac{n(A)}{n(S)}}$

${p(A)\:=\:\frac{9}{100}}$

${\boxed{p(A)\:=\:\frac{9}{100}\:or\:0.09}}$

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2. a prime number : €vent B

pri.m.e num.bers are = 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

${n(B)\:=\:25}$

${p(B)\:=\:\frac{n(B)}{n(S)}}$

${p(B)\:=\:\frac{25}{100}}$

${\boxed{p(B)\:=\:\frac{1}{4}\:or\:0.25}}$

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3. a two digit number : €vent C

two digit number betw.een 1 - 100 are 10,11,12,13,14...99

${n(C)\:=\:90}$

${p(C)\:=\:\frac{n(C)}{n(S)}}$

${p(C)\:=\:\frac{90}{100}}$

${\boxed{p(C)\:=\:\frac{9}{10}\:or\:0.9}}$

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4. a multiple of 9 : €vent D

Multiple of 9 = 9 , 18 , 27, 36, 45, 54, 63, 72, 81, 90 , 99

${n(D)\:=\:11}$

${p(D)\:=\:\frac{n(D)}{n(S)}}$

${p(D)\:=\:\frac{11}{100}}$

${\boxed{p(D)\:=\:\frac{11}{100}\:or\:0.11}}$