Probability ncert solutions Chapter 15 Exercise 15.1 Question 18
18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5
Solution:
(i)
Let E be the event of drawing a disc bearing two-digit number.
Number of possible outcomes favorable to event E=81
(There are 81 two digit numbers from 1 to 90)
Total number of possible outcomes=90 (There are 90 discs)
P(E)=(Number of possible outcomes favorable to event E)/(Total number of possible outcomes)
(ii)
Let A be the event of drawing a disc bearing a perfect square number.
The perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
Therefore, there are 9 perfect square numbers from 1 to 90.
Number of possible outcomes favorable to event A=9
(There are 9 perfect square numbers from 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(A)=(Number of possible outcomes favorable to event A)/(Total number of possible outcomes)
(iii)
Let B be the event of drawing a disc bearing a number divisible by 5.
The numbers divisible by 5 from numbers 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.
Therefore, there are 18 numbers divisible by 5 from numbers 1 to 90.
Number of outcomes favorable to event B=18
(There are 18 numbers divisible by 5 from numbers 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(B)=(Number of possible outcomes favorable to event B)/(Total number of possible outcomes)
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