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A bag contains cards which are numbered from $ 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
VerifiedVerified
420.6k+ views
Hint: Probability can be defined as the event which is likely to occur. It can be expressed as the ratio of favorable outcomes to the total possible outcomes. Here we will first find the total number of possible outcomes and then we will find the favorable outcomes for both the cases and simplify for the required probability.

Complete step by step solution:
The total possible outcomes from $ 2 $ to $ 90 = 89 $
 $ n(s) = 89 $
Let us assume that the A be an event of drawing a two digit number from the cards numbered from $ 2 $ to $ 90 $
Two digits number from the given range $ 2 $ to $ 90 = 10,11,12,....90 $
 $ n(A) = 81 $
Probability, $ P(A) = \dfrac{{Number\,{\text{of favourable outcomes}}}}{{Total{\text{ number of possible outcomes}}}} $
Place the values in the above equation –
 $ P(A) = \dfrac{{81}}{{89}} $
Let us assume that B be the event a number which is a perfect square
Perfect square from the given range $ 2 $ to $ 90 = 4,9,16,25,36,49,64,81 $
 $ n(B) = 8 $
Place the values in the probability formula –
 $ P(B) = \dfrac{8}{{89}} $

Note: Be good in finding the square numbers, square numbers are the numbers in which we get multiplying the same integers or numbers twice. Remember probability ranges between the values ranging between zero and one. The probability of an impossible event is zero and the probability of the sure event is one.