Question

A bag contains cards which are numbered form 2 to 90. A card is drawn at random from the bag. Find the probability that it bears
(i)a two digit number
(ii)a number which is a perfect square.

Answer 1

SOLUTION :  

Given : Cards marked with numbers from 2 to 90

Total number of outcomes = 90 - 2 + 1 =  89 ( both 2 & 90 are included)

(i)Let E1 = Event of getting a two digit number

Numbers which are 2 digits are from 10 to 90 = 90 - 10 + 1 = 81 ( both 10 & 90 are included)

Number of outcome favourable to E1 = 81

Probability (E1) = Number of favourable outcomes / Total number of outcomes

P(E1) = 81/89

Hence, the required probability of getting two digit number, P(E1) = 81/89  .

(ii)Let E2 = Event of getting a number which are perfect squares  

Numbers which are perfect squares = 4, 9, 16, 25 ,35,49,64, 81

Number of perfect squares = 8  

Number of outcome favourable to E2 = 8

Probability (E2) = Number of favourable outcomes / Total number of outcomes

P(E2) = 8/89  

Hence, the required probability of getting a number which are perfect squares ,P(E2) = 8/89 .

HOPE THIS ANSWER WILL HELP YOU ...

Answer 2

Total no. of possible outcomes = 89 {2, 3, 4, …., 90}

(i) Let E event of getting a 2 digit no.

No. favourable outcomes = 81 {10, 11, 12, 13, ….., 80}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 81/89

(ii) E event of getting a no. which is perfect square

No. favourable outcomes = 8 {4, 9, 16, 25, 36, 49, 64, 81}

P(E) = 8/89

hope this helps you out!