Question

One is asked to say a two digit number.What is the probability of being the numbers not a perfect square??​

Answer 1

Step-by-step explanation:

Two digits no.s = 10-99

n(S) = 90

Let A be the event of getting perfect squares.

No. of perfect squares in two digit numbers = 6 (16, 25, 36, 49, 64, 81)

n(A) = 6

$P(A) = \frac{n(A) }{n(S)} = \frac{6}{90} = \frac{1}{15}$

Let P(A') be the event of not getting perfect Square numbers.

$P(A') =1 - P(A)$

$= 1 - \frac{1}{15} = \frac{14}{15}$

Answer 2

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